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Номер 8.11, страница 58 - гдз по алгебре 10 класс учебник Абылкасымова, Жумагулова
Авторы: Абылкасымова А. Е., Жумагулова З. А.
Тип: Учебник
Издательство: Мектеп
Год издания: 2019 - 2026
ISBN: 978-601-07-1142-6
Утверждено Министерством образования и науки Республики Казахстан
Популярные ГДЗ в 10 классе
Глава 3. Тригонометрические уравнения и неравенства. Параграф 8. Решение тригонометрических уравнений - номер 8.11, страница 58.
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[] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_author" => Illuminate\Database\Eloquent\Collection {#1056 #items: array:2 [ 0 => App\Models\Term {#1057 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:12 [ "id" => 6997 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Абылкасымова" "field_bio" => null "field_degree_rp" => null "field_foreign_lang_name" => null "field_foreign_lang_patronymic" => null "field_foreign_lang_surname" => null "field_name" => "Алма" "field_patronymic" => "Есимбековна" 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"*" ] } ] #escapeWhenCastingToString: false } "branch_parent" => Illuminate\Database\Eloquent\Collection {#1076 #items: [] #escapeWhenCastingToString: false } ] #original: array:24 [ "id" => 1113562 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Тригонометрические уравнения и неравенства" "field_branch_order" => "3" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1050} "field_page_start" => "48" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1078} "branch_parent" => Illuminate\Database\Eloquent\Collection {#1076} ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "parent_branches" => Illuminate\Database\Eloquent\Collection {#1105 #items: array:2 [ 0 => App\Models\Branch {#1113 #connection: "mysql" #table: "branches" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 1113562 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Тригонометрические уравнения и неравенства" "field_branch_order" => "3" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1114 #items: array:1 [ 0 => App\Models\Term {#1047} ] #escapeWhenCastingToString: false } "field_page_start" => "48" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1115 #items: array:1 [ 0 => App\Models\Book {#1049} ] #escapeWhenCastingToString: false } "branch_parent" => Illuminate\Database\Eloquent\Collection {#1116 #items: [] #escapeWhenCastingToString: false } ] #original: array:24 [ "id" => 1113562 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Тригонометрические уравнения и неравенства" "field_branch_order" => "3" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1114} "field_page_start" => "48" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1115} "branch_parent" => Illuminate\Database\Eloquent\Collection {#1116} ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 1 => App\Models\Branch {#1117 #connection: "mysql" #table: "branches" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 1113564 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Решение тригонометрических уравнений" "field_branch_order" => "8" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1121 #items: array:1 [ 0 => App\Models\Term {#1118 #connection: "mysql" #table: "terms" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:7 [ "id" => 32 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Параграф" "field_cases" => array:6 [ …6] "field_page_check_not_needed" => null "field_short_name" => "§" ] #original: array:7 [ "id" => 32 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "name" => "Параграф" "field_cases" => array:6 [ …6] "field_page_check_not_needed" => null "field_short_name" => "§" ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "field_page_start" => "55" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1120 #items: array:1 [ 0 => App\Models\Book {#1049} ] #escapeWhenCastingToString: false } "branch_parent" => Illuminate\Database\Eloquent\Collection {#1119 #items: array:1 [ 0 => App\Models\Branch {#1045} ] #escapeWhenCastingToString: false } ] #original: array:24 [ "id" => 1113564 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_display_title" => "Решение тригонометрических уравнений" "field_branch_order" => "8" "field_url" => null "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1121} "field_page_start" => "55" "field_branch_display" => "0" "field_branch_expanded" => "0" "field_display_branch_in_title" => "1" "field_display_task_interval" => "0" "field_display_branch_page" => "1" "field_branch_title_in_content" => "0" "field_navigation_title" => null "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_branch_cover" => [] "field_branch_covers" => [] "book" => Illuminate\Database\Eloquent\Collection {#1120} "branch_parent" => Illuminate\Database\Eloquent\Collection {#1119} ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "content" => Illuminate\Database\Eloquent\Collection {#1102 #items: array:3 [ 0 => App\Models\Element {#1095 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:7 [ "id" => 1290764 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1087 #items: array:1 [ 0 => App\Models\Edition {#1094 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 4958 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Условие" "field_order" => "1" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1092 …2} "field_content_type" => "free" "field_content_mode" => "text, image" "field_page_content_mode" => "image" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Автор" "field_moderator" => "pasha" "field_edition_type" => "statement" "field_root_dir" => "0-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1090 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1091 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1088 …2} "field_content_source" => null ] #original: array:21 [ "id" => 4958 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Условие" "field_order" => "1" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1092 …2} "field_content_type" => "free" "field_content_mode" => "text, image" "field_page_content_mode" => "image" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Автор" "field_moderator" => "pasha" "field_edition_type" => "statement" "field_root_dir" => "0-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1090 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1091 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1088 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "1113754" "type" => "task" ] "text" => "<p><strong>8.11. a)</strong> $6\sin^2x = 4 + \sin 2x;$</p><p><strong>б)</strong> $3\sin 2x + 8\cos^2x = 7.$</p>" "img" => array:1 [ 0 => array:5 [ "name" => "11-1.jpg" "alt" => null "width" => "2429" "height" => 232 "path" => "/media/algebra_10/Abylkasymova-u/0-08/11-1.webp?ts=1753263047" ] ] ] #original: array:7 [ "id" => 1290764 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1087} "task" => array:2 [ "refs" => "1113754" "type" => "task" ] "text" => "<p><strong>8.11. a)</strong> $6\sin^2x = 4 + \sin 2x;$</p><p><strong>б)</strong> $3\sin 2x + 8\cos^2x = 7.$</p>" "img" => array:1 [ 0 => array:5 [ "name" => "11-1.jpg" "alt" => null "width" => "2429" "height" => 232 "path" => "/media/algebra_10/Abylkasymova-u/0-08/11-1.webp?ts=1753263047" ] ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 1 => App\Models\Element {#1089 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 1291278 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1079 #items: array:1 [ 0 => App\Models\Edition {#1086 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 4959 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение" "field_order" => "2" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1082 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "pasha" "field_edition_type" => "solution" "field_root_dir" => "1-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1084 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1083 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1080 …2} "field_content_source" => null ] #original: array:21 [ "id" => 4959 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение" "field_order" => "2" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1082 …2} "field_content_type" => "free" "field_content_mode" => "image" "field_page_content_mode" => "" "field_content_text_checked" => "0" "field_page_content_text_checked" => "0" "field_solution_author" => "Неизвестный" "field_moderator" => "pasha" "field_edition_type" => "solution" "field_root_dir" => "1-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1084 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1083 …2} "field_process_formula" => "" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1080 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "1113754" "type" => "task" ] "img" => array:2 [ 0 => array:5 [ "name" => "11-1.jpg" "alt" => null "width" => "1753" "height" => 2118 "path" => "/media/algebra_10/Abylkasymova-u/1-08/11-1.webp?ts=1753263231" ] 1 => array:5 [ "name" => "11-2.jpg" "alt" => null "width" => "1828" "height" => 481 "path" => "/media/algebra_10/Abylkasymova-u/1-08/11-2.webp?ts=1753263231" ] ] ] #original: array:6 [ "id" => 1291278 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1079} "task" => array:2 [ "refs" => "1113754" "type" => "task" ] "img" => array:2 [ 0 => array:5 [ "name" => "11-1.jpg" "alt" => null "width" => "1753" "height" => 2118 "path" => "/media/algebra_10/Abylkasymova-u/1-08/11-1.webp?ts=1753263231" ] 1 => array:5 [ "name" => "11-2.jpg" "alt" => null "width" => "1828" "height" => 481 "path" => "/media/algebra_10/Abylkasymova-u/1-08/11-2.webp?ts=1753263231" ] ] ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 2 => App\Models\Element {#1081 #connection: "mysql" #table: "elements" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:6 [ "id" => 1549407 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1108 #items: array:1 [ 0 => App\Models\Edition {#1077 #connection: "mysql" #table: "editions" #primaryKey: "id" #keyType: "int" +incrementing: false #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:21 [ "id" => 5655 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 2" "field_order" => "3" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1041 …2} "field_content_type" => "free" "field_content_mode" => "text" "field_page_content_mode" => "" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Gemini 2.5 Pro" "field_moderator" => "tolik" "field_edition_type" => "solution" "field_root_dir" => "2-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1034 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1039 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1104 …2} "field_content_source" => null ] #original: array:21 [ "id" => 5655 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "title" => "Решение 2" "field_order" => "3" "field_publisher" => Illuminate\Database\Eloquent\Collection {#1041 …2} "field_content_type" => "free" "field_content_mode" => "text" "field_page_content_mode" => "" "field_content_text_checked" => null "field_page_content_text_checked" => "0" "field_solution_author" => "Gemini 2.5 Pro" "field_moderator" => "tolik" "field_edition_type" => "solution" "field_root_dir" => "2-" "field_responsible" => Illuminate\Database\Eloquent\Collection {#1034 …2} "field_comment" => null "field_similar_book" => Illuminate\Database\Eloquent\Collection {#1039 …2} "field_process_formula" => "katex" "field_edition_group" => Illuminate\Database\Eloquent\Collection {#1104 …2} "field_content_source" => null ] #changes: [] #casts: [] #classCastCache: [] #attributeCastCache: [] #dateFormat: null #appends: [] #dispatchesEvents: [] #observables: [] #relations: [] #touches: [] +timestamps: true +usesUniqueIds: false #hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } ] #escapeWhenCastingToString: false } "task" => array:2 [ "refs" => "1113754" "type" => "task" ] "text" => "<p><strong>a)</strong> $6\sin^2x = 4 + \sin 2x$</p><p>Для решения данного тригонометрического уравнения воспользуемся формулой двойного угла $\sin 2x = 2\sin x \cos x$ и основным тригонометрическим тождеством, представив число 4 как $4 \cdot 1 = 4(\sin^2x + \cos^2x)$.</p><p>Перепишем уравнение:</p><p>$6\sin^2x = 4(\sin^2x + \cos^2x) + 2\sin x \cos x$</p><p>Раскроем скобки и перенесем все члены в левую часть:</p><p>$6\sin^2x - 4\sin^2x - 4\cos^2x - 2\sin x \cos x = 0$</p><p>$2\sin^2x - 2\sin x \cos x - 4\cos^2x = 0$</p><p>Разделим обе части уравнения на 2:</p><p>$\sin^2x - \sin x \cos x - 2\cos^2x = 0$</p><p>Это однородное тригонометрическое уравнение второй степени. Проверим, является ли $\cos x = 0$ решением. Если $\cos x = 0$, то $\sin^2x = 1$. Подставив в уравнение, получим $1 - 0 - 0 = 0$, что неверно. Следовательно, $\cos x \neq 0$, и мы можем разделить обе части уравнения на $\cos^2x$.</p><p>$\frac{\sin^2x}{\cos^2x} - \frac{\sin x \cos x}{\cos^2x} - \frac{2\cos^2x}{\cos^2x} = 0$</p><p>$\tan^2x - \tan x - 2 = 0$</p><p>Сделаем замену переменной $t = \tan x$. Уравнение примет вид квадратного уравнения:</p><p>$t^2 - t - 2 = 0$</p><p>Решим это уравнение. По теореме Виета или через дискриминант находим корни:</p><p>$D = (-1)^2 - 4(1)(-2) = 1 + 8 = 9$</p><p>$t_1 = \frac{1 + \sqrt{9}}{2} = \frac{1+3}{2} = 2$</p><p>$t_2 = \frac{1 - \sqrt{9}}{2} = \frac{1-3}{2} = -1$</p><p>Теперь вернемся к исходной переменной $x$.</p><p><strong>1.</strong> Если $\tan x = 2$, то $x = \arctan 2 + \pi n, n \in \mathbb{Z}$.</p><p><strong>2.</strong> Если $\tan x = -1$, то $x = \arctan(-1) + \pi k = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}$.</p><p><strong>Ответ:</strong> $x = \arctan 2 + \pi n, n \in \mathbb{Z}$; $x = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}$.</p><p><strong>б)</strong> $3\sin 2x + 8\cos^2x = 7$</p><p>Используем формулу двойного угла $\sin 2x = 2\sin x \cos x$ и основное тригонометрическое тождество, представив 7 как $7 \cdot 1 = 7(\sin^2x + \cos^2x)$.</p><p>Подставим эти выражения в исходное уравнение:</p><p>$3(2\sin x \cos x) + 8\cos^2x = 7(\sin^2x + \cos^2x)$</p><p>$6\sin x \cos x + 8\cos^2x = 7\sin^2x + 7\cos^2x$</p><p>Перенесем все члены в одну сторону, чтобы получить однородное уравнение:</p><p>$7\sin^2x - 6\sin x \cos x + 7\cos^2x - 8\cos^2x = 0$</p><p>$7\sin^2x - 6\sin x \cos x - \cos^2x = 0$</p><p>Это однородное тригонометрическое уравнение второй степени. Убедимся, что $\cos x \neq 0$. Если $\cos x = 0$, то $\sin^2x = 1$. Уравнение примет вид $7(1) - 0 - 0 = 0$, что является неверным равенством ($7=0$). Значит, можно разделить обе части на $\cos^2x$.</p><p>$7\frac{\sin^2x}{\cos^2x} - 6\frac{\sin x \cos x}{\cos^2x} - \frac{\cos^2x}{\cos^2x} = 0$</p><p>$7\tan^2x - 6\tan x - 1 = 0$</p><p>Произведем замену $t = \tan x$:</p><p>$7t^2 - 6t - 1 = 0$</p><p>Решим полученное квадратное уравнение с помощью дискриминанта:</p><p>$D = (-6)^2 - 4(7)(-1) = 36 + 28 = 64$</p><p>$t_1 = \frac{6 + \sqrt{64}}{2 \cdot 7} = \frac{6+8}{14} = \frac{14}{14} = 1$</p><p>$t_2 = \frac{6 - \sqrt{64}}{2 \cdot 7} = \frac{6-8}{14} = \frac{-2}{14} = -\frac{1}{7}$</p><p>Выполним обратную замену.</p><p><strong>1.</strong> Если $\tan x = 1$, то $x = \arctan 1 + \pi n = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$.</p><p><strong>2.</strong> Если $\tan x = -\frac{1}{7}$, то $x = \arctan(-\frac{1}{7}) + \pi k = -\arctan\frac{1}{7} + \pi k, k \in \mathbb{Z}$.</p><p><strong>Ответ:</strong> $x = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$; $x = -\arctan\frac{1}{7} + \pi k, k \in \mathbb{Z}$.</p>" ] #original: array:6 [ "id" => 1549407 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "edition" => Illuminate\Database\Eloquent\Collection {#1108} "task" => array:2 [ "refs" => "1113754" "type" => "task" ] "text" => "<p><strong>a)</strong> $6\sin^2x = 4 + \sin 2x$</p><p>Для решения данного тригонометрического уравнения воспользуемся формулой двойного угла $\sin 2x = 2\sin x \cos x$ и основным тригонометрическим тождеством, представив число 4 как $4 \cdot 1 = 4(\sin^2x + \cos^2x)$.</p><p>Перепишем уравнение:</p><p>$6\sin^2x = 4(\sin^2x + \cos^2x) + 2\sin x \cos x$</p><p>Раскроем скобки и перенесем все члены в левую часть:</p><p>$6\sin^2x - 4\sin^2x - 4\cos^2x - 2\sin x \cos x = 0$</p><p>$2\sin^2x - 2\sin x \cos x - 4\cos^2x = 0$</p><p>Разделим обе части уравнения на 2:</p><p>$\sin^2x - \sin x \cos x - 2\cos^2x = 0$</p><p>Это однородное тригонометрическое уравнение второй степени. Проверим, является ли $\cos x = 0$ решением. Если $\cos x = 0$, то $\sin^2x = 1$. Подставив в уравнение, получим $1 - 0 - 0 = 0$, что неверно. Следовательно, $\cos x \neq 0$, и мы можем разделить обе части уравнения на $\cos^2x$.</p><p>$\frac{\sin^2x}{\cos^2x} - \frac{\sin x \cos x}{\cos^2x} - \frac{2\cos^2x}{\cos^2x} = 0$</p><p>$\tan^2x - \tan x - 2 = 0$</p><p>Сделаем замену переменной $t = \tan x$. Уравнение примет вид квадратного уравнения:</p><p>$t^2 - t - 2 = 0$</p><p>Решим это уравнение. По теореме Виета или через дискриминант находим корни:</p><p>$D = (-1)^2 - 4(1)(-2) = 1 + 8 = 9$</p><p>$t_1 = \frac{1 + \sqrt{9}}{2} = \frac{1+3}{2} = 2$</p><p>$t_2 = \frac{1 - \sqrt{9}}{2} = \frac{1-3}{2} = -1$</p><p>Теперь вернемся к исходной переменной $x$.</p><p><strong>1.</strong> Если $\tan x = 2$, то $x = \arctan 2 + \pi n, n \in \mathbb{Z}$.</p><p><strong>2.</strong> Если $\tan x = -1$, то $x = \arctan(-1) + \pi k = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}$.</p><p><strong>Ответ:</strong> $x = \arctan 2 + \pi n, n \in \mathbb{Z}$; $x = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}$.</p><p><strong>б)</strong> $3\sin 2x + 8\cos^2x = 7$</p><p>Используем формулу двойного угла $\sin 2x = 2\sin x \cos x$ и основное тригонометрическое тождество, представив 7 как $7 \cdot 1 = 7(\sin^2x + \cos^2x)$.</p><p>Подставим эти выражения в исходное уравнение:</p><p>$3(2\sin x \cos x) + 8\cos^2x = 7(\sin^2x + \cos^2x)$</p><p>$6\sin x \cos x + 8\cos^2x = 7\sin^2x + 7\cos^2x$</p><p>Перенесем все члены в одну сторону, чтобы получить однородное уравнение:</p><p>$7\sin^2x - 6\sin x \cos x + 7\cos^2x - 8\cos^2x = 0$</p><p>$7\sin^2x - 6\sin x \cos x - \cos^2x = 0$</p><p>Это однородное тригонометрическое уравнение второй степени. Убедимся, что $\cos x \neq 0$. Если $\cos x = 0$, то $\sin^2x = 1$. Уравнение примет вид $7(1) - 0 - 0 = 0$, что является неверным равенством ($7=0$). Значит, можно разделить обе части на $\cos^2x$.</p><p>$7\frac{\sin^2x}{\cos^2x} - 6\frac{\sin x \cos x}{\cos^2x} - \frac{\cos^2x}{\cos^2x} = 0$</p><p>$7\tan^2x - 6\tan x - 1 = 0$</p><p>Произведем замену $t = \tan x$:</p><p>$7t^2 - 6t - 1 = 0$</p><p>Решим полученное квадратное уравнение с помощью дискриминанта:</p><p>$D = (-6)^2 - 4(7)(-1) = 36 + 28 = 64$</p><p>$t_1 = \frac{6 + \sqrt{64}}{2 \cdot 7} = \frac{6+8}{14} = \frac{14}{14} = 1$</p><p>$t_2 = \frac{6 - \sqrt{64}}{2 \cdot 7} = \frac{6-8}{14} = \frac{-2}{14} = -\frac{1}{7}$</p><p>Выполним обратную замену.</p><p><strong>1.</strong> Если $\tan x = 1$, то $x = \arctan 1 + \pi n = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$.</p><p><strong>2.</strong> Если $\tan x = -\frac{1}{7}$, то $x = \arctan(-\frac{1}{7}) + \pi k = -\arctan\frac{1}{7} + \pi k, k \in \mathbb{Z}$.</p><p><strong>Ответ:</strong> $x = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$; $x = -\arctan\frac{1}{7} + \pi k, k \in \mathbb{Z}$.</p>" ] #changes: [] #casts: [] 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#hidden: [] #visible: [] #fillable: [] #guarded: array:1 [ 0 => "*" ] } 4 => App\Models\Task {#1256 #connection: "mysql" #table: "tasks" #primaryKey: "id" #keyType: "int" +incrementing: true #with: [] #withCount: [] +preventsLazyLoading: false #perPage: 15 +exists: true +wasRecentlyCreated: false #escapeWhenCastingToString: false #attributes: array:24 [ "id" => 1113747 "created_at" => "2026-04-10 13:58:26" "updated_at" => null "field_page_start" => "58" "field_page_end" => null "field_url" => "/10-klass/algebra/abylkasymova-uchebnik/08-4" "field_display_title" => "8.4" "field_outside_task" => null "field_task_type" => Illuminate\Database\Eloquent\Collection {#1253 …2} "field_metatags_title" => null "field_metatags_description" => null "field_h1" => null "field_description_top" => null "field_description_bottom" => null "field_match" => null "breadcrumbs" => [] "edition_groups" => Illuminate\Database\Eloquent\Collection {#1271 …2} "top_parent_branch" => 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№8.11 (с. 58)
Условие. №8.11 (с. 58)
скриншот условия
8.11. a) $6\sin^2x = 4 + \sin 2x;$
б) $3\sin 2x + 8\cos^2x = 7.$
Решение. №8.11 (с. 58)
Решение 2. №8.11 (с. 58)
a) $6\sin^2x = 4 + \sin 2x$
Для решения данного тригонометрического уравнения воспользуемся формулой двойного угла $\sin 2x = 2\sin x \cos x$ и основным тригонометрическим тождеством, представив число 4 как $4 \cdot 1 = 4(\sin^2x + \cos^2x)$.
Перепишем уравнение:
$6\sin^2x = 4(\sin^2x + \cos^2x) + 2\sin x \cos x$
Раскроем скобки и перенесем все члены в левую часть:
$6\sin^2x - 4\sin^2x - 4\cos^2x - 2\sin x \cos x = 0$
$2\sin^2x - 2\sin x \cos x - 4\cos^2x = 0$
Разделим обе части уравнения на 2:
$\sin^2x - \sin x \cos x - 2\cos^2x = 0$
Это однородное тригонометрическое уравнение второй степени. Проверим, является ли $\cos x = 0$ решением. Если $\cos x = 0$, то $\sin^2x = 1$. Подставив в уравнение, получим $1 - 0 - 0 = 0$, что неверно. Следовательно, $\cos x \neq 0$, и мы можем разделить обе части уравнения на $\cos^2x$.
$\frac{\sin^2x}{\cos^2x} - \frac{\sin x \cos x}{\cos^2x} - \frac{2\cos^2x}{\cos^2x} = 0$
$\tan^2x - \tan x - 2 = 0$
Сделаем замену переменной $t = \tan x$. Уравнение примет вид квадратного уравнения:
$t^2 - t - 2 = 0$
Решим это уравнение. По теореме Виета или через дискриминант находим корни:
$D = (-1)^2 - 4(1)(-2) = 1 + 8 = 9$
$t_1 = \frac{1 + \sqrt{9}}{2} = \frac{1+3}{2} = 2$
$t_2 = \frac{1 - \sqrt{9}}{2} = \frac{1-3}{2} = -1$
Теперь вернемся к исходной переменной $x$.
1. Если $\tan x = 2$, то $x = \arctan 2 + \pi n, n \in \mathbb{Z}$.
2. Если $\tan x = -1$, то $x = \arctan(-1) + \pi k = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}$.
Ответ: $x = \arctan 2 + \pi n, n \in \mathbb{Z}$; $x = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}$.
б) $3\sin 2x + 8\cos^2x = 7$
Используем формулу двойного угла $\sin 2x = 2\sin x \cos x$ и основное тригонометрическое тождество, представив 7 как $7 \cdot 1 = 7(\sin^2x + \cos^2x)$.
Подставим эти выражения в исходное уравнение:
$3(2\sin x \cos x) + 8\cos^2x = 7(\sin^2x + \cos^2x)$
$6\sin x \cos x + 8\cos^2x = 7\sin^2x + 7\cos^2x$
Перенесем все члены в одну сторону, чтобы получить однородное уравнение:
$7\sin^2x - 6\sin x \cos x + 7\cos^2x - 8\cos^2x = 0$
$7\sin^2x - 6\sin x \cos x - \cos^2x = 0$
Это однородное тригонометрическое уравнение второй степени. Убедимся, что $\cos x \neq 0$. Если $\cos x = 0$, то $\sin^2x = 1$. Уравнение примет вид $7(1) - 0 - 0 = 0$, что является неверным равенством ($7=0$). Значит, можно разделить обе части на $\cos^2x$.
$7\frac{\sin^2x}{\cos^2x} - 6\frac{\sin x \cos x}{\cos^2x} - \frac{\cos^2x}{\cos^2x} = 0$
$7\tan^2x - 6\tan x - 1 = 0$
Произведем замену $t = \tan x$:
$7t^2 - 6t - 1 = 0$
Решим полученное квадратное уравнение с помощью дискриминанта:
$D = (-6)^2 - 4(7)(-1) = 36 + 28 = 64$
$t_1 = \frac{6 + \sqrt{64}}{2 \cdot 7} = \frac{6+8}{14} = \frac{14}{14} = 1$
$t_2 = \frac{6 - \sqrt{64}}{2 \cdot 7} = \frac{6-8}{14} = \frac{-2}{14} = -\frac{1}{7}$
Выполним обратную замену.
1. Если $\tan x = 1$, то $x = \arctan 1 + \pi n = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$.
2. Если $\tan x = -\frac{1}{7}$, то $x = \arctan(-\frac{1}{7}) + \pi k = -\arctan\frac{1}{7} + \pi k, k \in \mathbb{Z}$.
Ответ: $x = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$; $x = -\arctan\frac{1}{7} + \pi k, k \in \mathbb{Z}$.
Другие задания:
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