Номер 28, страница 48, часть 2 - гдз по алгебре 10 класс учебник Пак, Ардакулы

Авторы: Пак О. В., Ардакулы Д., Ескендирова Е. В.

Тип: Учебник

Издательство: Алматыкітап баспасы

Год издания: 2019 - 2026

Часть: 2

ISBN: 978-601-01-3958-9

Рекомендовано Министерством образования и науки Республики Казахстан

Часть 2. Глава 5. Производная. Параграф 2. Понятие производной. 2.5. Формулы для нахождения производной. Задачи - номер 28, страница 48.

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            "branch_parent" => Illuminate\Database\Eloquent\Collection {#1095
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                    "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1097 …2}
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        }
        2 => App\Models\Branch {#1101
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          #table: "branches"
          #primaryKey: "id"
          #keyType: "int"
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                0 => App\Models\Term {#1103
                  #connection: "mysql"
                  #table: "terms"
                  #primaryKey: "id"
                  #keyType: "int"
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                  #attributes: array:7 [
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                    "created_at" => "2026-04-10 13:58:26"
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                    "field_page_check_not_needed" => null
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              ]
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            "book" => Illuminate\Database\Eloquent\Collection {#1104
              #items: array:1 [
                0 => App\Models\Book {#1105
                  #connection: "mysql"
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                  #primaryKey: "id"
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                    "field_subject" => Illuminate\Database\Eloquent\Collection {#1106 …2}
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                    "field_cover" => array:1 [ …1]
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                0 => App\Models\Branch {#1134
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                    "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1135 …2}
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                    "book" => Illuminate\Database\Eloquent\Collection {#1137 …2}
                    "branch_parent" => Illuminate\Database\Eloquent\Collection {#1138 …2}
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            0 => "*"
          ]
        }
        3 => App\Models\Branch {#1172
          #connection: "mysql"
          #table: "branches"
          #primaryKey: "id"
          #keyType: "int"
          +incrementing: true
          #with: []
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            "created_at" => "2026-04-10 13:58:26"
            "updated_at" => null
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            "field_branch_order" => null
            "field_url" => null
            "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1173
              #items: []
              #escapeWhenCastingToString: false
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            "field_page_start" => "37"
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            "id" => 1300909
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            "edition" => Illuminate\Database\Eloquent\Collection {#1384
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            "text" => "<p><strong>28.</strong> При каких значениях переменной значение функции $g(x)$ равно значению производной функции $f(x)$, если:</p><p><strong>а)</strong> $f(x) = -\frac{2\cos5x}{5} + \cos\frac{\pi}{3}$, $g(x) = \sin3x + \sin7x$;</p><p><strong>б)</strong> $f(x) = 2x + \cos^2\frac{4\pi}{17}$, $g(x) = \sin^2 2x + \sin^2 3x + \sin^2 4x + \sin^2 5x$;</p><p><strong>в)</strong> $f(x) = \frac{\cos8x}{8} - \frac{\cos2x}{2}$, $g(x) = 2\sin3x \cos x$;</p><p><strong>г)</strong> $f(x) = \frac{\sin4x}{8} + \frac{3}{4}x + \frac{\sin2\pi}{21}$, $g(x) = \sin^4 x + \cos^4 x?$</p>"
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            "task" => array:2 [
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              "type" => "task"
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            "text" => "<p><strong>28.</strong> При каких значениях переменной значение функции $g(x)$ равно значению производной функции $f(x)$, если:</p><p><strong>а)</strong> $f(x) = -\frac{2\cos5x}{5} + \cos\frac{\pi}{3}$, $g(x) = \sin3x + \sin7x$;</p><p><strong>б)</strong> $f(x) = 2x + \cos^2\frac{4\pi}{17}$, $g(x) = \sin^2 2x + \sin^2 3x + \sin^2 4x + \sin^2 5x$;</p><p><strong>в)</strong> $f(x) = \frac{\cos8x}{8} - \frac{\cos2x}{2}$, $g(x) = 2\sin3x \cos x$;</p><p><strong>г)</strong> $f(x) = \frac{\sin4x}{8} + \frac{3}{4}x + \frac{\sin2\pi}{21}$, $g(x) = \sin^4 x + \cos^4 x?$</p>"
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            "created_at" => "2026-04-10 13:58:26"
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            "edition" => Illuminate\Database\Eloquent\Collection {#1392
              #items: array:1 [
                0 => App\Models\Edition {#1393
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            "text" => "<p><strong>а)</strong></p> <p>Сначала найдем производную функции $f(x) = -\frac{2\cos{5x}}{5} + \cos{\frac{\pi}{3}}$.</p> <p>Поскольку $\cos{\frac{\pi}{3}}$ является константой, ее производная равна нулю. Производная от $\cos{5x}$ равна $-5\sin{5x}$.</p> <p>$f'(x) = \left(-\frac{2}{5}\cos{5x} + \cos{\frac{\pi}{3}}\right)' = -\frac{2}{5}(-\sin{5x} \cdot 5) + 0 = 2\sin{5x}$.</p> <p>Теперь приравняем значение функции $g(x)$ к значению производной $f'(x)$:</p> <p>$g(x) = f'(x)$</p> <p>$\sin{3x} + \sin{7x} = 2\sin{5x}$</p> <p>Используем формулу суммы синусов: $\sin{\alpha} + \sin{\beta} = 2\sin{\frac{\alpha+\beta}{2}}\cos{\frac{\alpha-\beta}{2}}$.</p> <p>$2\sin{\frac{3x+7x}{2}}\cos{\frac{7x-3x}{2}} = 2\sin{5x}$</p> <p>$2\sin{5x}\cos{2x} = 2\sin{5x}$</p> <p>$2\sin{5x}\cos{2x} - 2\sin{5x} = 0$</p> <p>$2\sin{5x}(\cos{2x} - 1) = 0$</p> <p>Это уравнение распадается на два:</p> <p>1) $\sin{5x} = 0$</p> <p>$5x = \pi k, \quad k \in Z$</p> <p>$x = \frac{\pi k}{5}, \quad k \in Z$</p> <p>2) $\cos{2x} - 1 = 0 \implies \cos{2x} = 1$</p> <p>$2x = 2\pi n, \quad n \in Z$</p> <p>$x = \pi n, \quad n \in Z$</p> <p>Заметим, что множество решений $x = \pi n$ является подмножеством множества решений $x = \frac{\pi k}{5}$ (при $k=5n$). Следовательно, общее решение - это $x = \frac{\pi k}{5}$.</p> <p><strong>Ответ:</strong> $x = \frac{\pi k}{5}, \quad k \in Z$.</p> <p><strong>б)</strong></p> <p>Найдем производную функции $f(x) = 2x + \cos^2{\frac{4\pi}{17}}$.</p> <p>Слагаемое $\cos^2{\frac{4\pi}{17}}$ является константой, поэтому его производная равна нулю.</p> <p>$f'(x) = (2x)' + (\cos^2{\frac{4\pi}{17}})' = 2 + 0 = 2$.</p> <p>Приравниваем $g(x)$ к $f'(x)$:</p> <p>$\sin^2{2x} + \sin^2{3x} + \sin^2{4x} + \sin^2{5x} = 2$</p> <p>Используем формулу понижения степени $\sin^2{\alpha} = \frac{1-\cos{2\alpha}}{2}$:</p> <p>$\frac{1-\cos{4x}}{2} + \frac{1-\cos{6x}}{2} + \frac{1-\cos{8x}}{2} + \frac{1-\cos{10x}}{2} = 2$</p> <p>$1-\cos{4x} + 1-\cos{6x} + 1-\cos{8x} + 1-\cos{10x} = 4$</p> <p>$4 - (\cos{4x} + \cos{6x} + \cos{8x} + \cos{10x}) = 4$</p> <p>$\cos{4x} + \cos{6x} + \cos{8x} + \cos{10x} = 0$</p> <p>Сгруппируем слагаемые и применим формулу суммы косинусов $\cos{\alpha} + \cos{\beta} = 2\cos{\frac{\alpha+\beta}{2}}\cos{\frac{\alpha-\beta}{2}}$:</p> <p>$(\cos{10x} + \cos{4x}) + (\cos{8x} + \cos{6x}) = 0$</p> <p>$2\cos{\frac{10x+4x}{2}}\cos{\frac{10x-4x}{2}} + 2\cos{\frac{8x+6x}{2}}\cos{\frac{8x-6x}{2}} = 0$</p> <p>$2\cos{7x}\cos{3x} + 2\cos{7x}\cos{x} = 0$</p> <p>$2\cos{7x}(\cos{3x} + \cos{x}) = 0$</p> <p>Снова применяем формулу суммы косинусов:</p> <p>$2\cos{7x}(2\cos{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}) = 0$</p> <p>$4\cos{7x}\cos{2x}\cos{x} = 0$</p> <p>Получаем три уравнения:</p> <p>1) $\cos{x} = 0 \implies x = \frac{\pi}{2} + \pi k, \quad k \in Z$</p> <p>2) $\cos{2x} = 0 \implies 2x = \frac{\pi}{2} + \pi n \implies x = \frac{\pi}{4} + \frac{\pi n}{2}, \quad n \in Z$</p> <p>3) $\cos{7x} = 0 \implies 7x = \frac{\pi}{2} + \pi m \implies x = \frac{\pi}{14} + \frac{\pi m}{7}, \quad m \in Z$</p> <p>Решения $x = \frac{\pi}{2} + \pi k$ являются подмножеством решений $x = \frac{\pi}{14} + \frac{\pi m}{7}$ (при $m = 3+7k$), поэтому первые можно не указывать в ответе.</p> <p><strong>Ответ:</strong> $x = \frac{\pi}{4} + \frac{\pi n}{2}, \quad n \in Z; \quad x = \frac{\pi}{14} + \frac{\pi m}{7}, \quad m \in Z$.</p> <p><strong>в)</strong></p> <p>Найдем производную функции $f(x) = -\frac{\cos{8x}}{8} - \frac{\cos{2x}}{2}$.</p> <p>$f'(x) = \left(-\frac{1}{8}\cos{8x}\right)' - \left(\frac{1}{2}\cos{2x}\right)' = -\frac{1}{8}(-\sin{8x} \cdot 8) - \frac{1}{2}(-\sin{2x} \cdot 2) = \sin{8x} + \sin{2x}$.</p> <p>Преобразуем функцию $g(x)$ по формуле произведения синуса на косинус $2\sin{\alpha}\cos{\beta} = \sin{(\alpha+\beta)} + \sin{(\alpha-\beta)}$:</p> <p>$g(x) = 2\sin{3x}\cos{x} = \sin(3x+x) + \sin(3x-x) = \sin{4x} + \sin{2x}$.</p> <p>Приравняем $g(x)$ к $f'(x)$:</p> <p>$\sin{4x} + \sin{2x} = \sin{8x} + \sin{2x}$</p> <p>$\sin{4x} = \sin{8x}$</p> <p>$\sin{8x} - \sin{4x} = 0$</p> <p>Используем формулу разности синусов $\sin{\alpha} - \sin{\beta} = 2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}$:</p> <p>$2\cos{\frac{8x+4x}{2}}\sin{\frac{8x-4x}{2}} = 0$</p> <p>$2\cos{6x}\sin{2x} = 0$</p> <p>Получаем два уравнения:</p> <p>1) $\sin{2x} = 0$</p> <p>$2x = \pi n \implies x = \frac{\pi n}{2}, \quad n \in Z$</p> <p>2) $\cos{6x} = 0$</p> <p>$6x = \frac{\pi}{2} + \pi k \implies x = \frac{\pi}{12} + \frac{\pi k}{6}, \quad k \in Z$</p> <p><strong>Ответ:</strong> $x = \frac{\pi n}{2}, \quad n \in Z; \quad x = \frac{\pi}{12} + \frac{\pi k}{6}, \quad k \in Z$.</p> <p><strong>г)</strong></p> <p>Найдем производную функции $f(x) = \frac{\sin{4x}}{8} + \frac{3}{4}x + \frac{\sin{2\pi}}{21}$.</p> <p>Слагаемое $\frac{\sin{2\pi}}{21}$ является константой (равно 0), его производная равна нулю.</p> <p>$f'(x) = \left(\frac{1}{8}\sin{4x}\right)' + \left(\frac{3}{4}x\right)' + 0 = \frac{1}{8}(\cos{4x} \cdot 4) + \frac{3}{4} = \frac{1}{2}\cos{4x} + \frac{3}{4}$.</p> <p>Теперь преобразуем функцию $g(x) = \sin^4{x} + \cos^4{x}$:</p> <p>$g(x) = (\sin^2{x} + \cos^2{x})^2 - 2\sin^2{x}\cos^2{x} = 1^2 - 2(\sin{x}\cos{x})^2 = 1 - 2\left(\frac{\sin{2x}}{2}\right)^2 = 1 - 2\frac{\sin^2{2x}}{4} = 1 - \frac{1}{2}\sin^2{2x}$.</p> <p>Используем формулу понижения степени $\sin^2{\alpha} = \frac{1-\cos{2\alpha}}{2}$:</p> <p>$g(x) = 1 - \frac{1}{2}\left(\frac{1-\cos{4x}}{2}\right) = 1 - \frac{1-\cos{4x}}{4} = \frac{4 - (1-\cos{4x})}{4} = \frac{3+\cos{4x}}{4} = \frac{3}{4} + \frac{1}{4}\cos{4x}$.</p> <p>Приравняем $g(x)$ к $f'(x)$:</p> <p>$\frac{3}{4} + \frac{1}{4}\cos{4x} = \frac{1}{2}\cos{4x} + \frac{3}{4}$</p> <p>$\frac{1}{4}\cos{4x} = \frac{1}{2}\cos{4x}$</p> <p>$\frac{1}{2}\cos{4x} - \frac{1}{4}\cos{4x} = 0$</p> <p>$\frac{1}{4}\cos{4x} = 0$</p> <p>$\cos{4x} = 0$</p> <p>$4x = \frac{\pi}{2} + \pi k, \quad k \in Z$</p> <p>$x = \frac{\pi}{8} + \frac{\pi k}{4}, \quad k \in Z$</p> <p><strong>Ответ:</strong> $x = \frac{\pi}{8} + \frac{\pi k}{4}, \quad k \in Z$.</p>"
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            "created_at" => "2026-04-10 13:58:26"
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            "edition" => Illuminate\Database\Eloquent\Collection {#1392}
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            "text" => "<p><strong>а)</strong></p> <p>Сначала найдем производную функции $f(x) = -\frac{2\cos{5x}}{5} + \cos{\frac{\pi}{3}}$.</p> <p>Поскольку $\cos{\frac{\pi}{3}}$ является константой, ее производная равна нулю. Производная от $\cos{5x}$ равна $-5\sin{5x}$.</p> <p>$f'(x) = \left(-\frac{2}{5}\cos{5x} + \cos{\frac{\pi}{3}}\right)' = -\frac{2}{5}(-\sin{5x} \cdot 5) + 0 = 2\sin{5x}$.</p> <p>Теперь приравняем значение функции $g(x)$ к значению производной $f'(x)$:</p> <p>$g(x) = f'(x)$</p> <p>$\sin{3x} + \sin{7x} = 2\sin{5x}$</p> <p>Используем формулу суммы синусов: $\sin{\alpha} + \sin{\beta} = 2\sin{\frac{\alpha+\beta}{2}}\cos{\frac{\alpha-\beta}{2}}$.</p> <p>$2\sin{\frac{3x+7x}{2}}\cos{\frac{7x-3x}{2}} = 2\sin{5x}$</p> <p>$2\sin{5x}\cos{2x} = 2\sin{5x}$</p> <p>$2\sin{5x}\cos{2x} - 2\sin{5x} = 0$</p> <p>$2\sin{5x}(\cos{2x} - 1) = 0$</p> <p>Это уравнение распадается на два:</p> <p>1) $\sin{5x} = 0$</p> <p>$5x = \pi k, \quad k \in Z$</p> <p>$x = \frac{\pi k}{5}, \quad k \in Z$</p> <p>2) $\cos{2x} - 1 = 0 \implies \cos{2x} = 1$</p> <p>$2x = 2\pi n, \quad n \in Z$</p> <p>$x = \pi n, \quad n \in Z$</p> <p>Заметим, что множество решений $x = \pi n$ является подмножеством множества решений $x = \frac{\pi k}{5}$ (при $k=5n$). Следовательно, общее решение - это $x = \frac{\pi k}{5}$.</p> <p><strong>Ответ:</strong> $x = \frac{\pi k}{5}, \quad k \in Z$.</p> <p><strong>б)</strong></p> <p>Найдем производную функции $f(x) = 2x + \cos^2{\frac{4\pi}{17}}$.</p> <p>Слагаемое $\cos^2{\frac{4\pi}{17}}$ является константой, поэтому его производная равна нулю.</p> <p>$f'(x) = (2x)' + (\cos^2{\frac{4\pi}{17}})' = 2 + 0 = 2$.</p> <p>Приравниваем $g(x)$ к $f'(x)$:</p> <p>$\sin^2{2x} + \sin^2{3x} + \sin^2{4x} + \sin^2{5x} = 2$</p> <p>Используем формулу понижения степени $\sin^2{\alpha} = \frac{1-\cos{2\alpha}}{2}$:</p> <p>$\frac{1-\cos{4x}}{2} + \frac{1-\cos{6x}}{2} + \frac{1-\cos{8x}}{2} + \frac{1-\cos{10x}}{2} = 2$</p> <p>$1-\cos{4x} + 1-\cos{6x} + 1-\cos{8x} + 1-\cos{10x} = 4$</p> <p>$4 - (\cos{4x} + \cos{6x} + \cos{8x} + \cos{10x}) = 4$</p> <p>$\cos{4x} + \cos{6x} + \cos{8x} + \cos{10x} = 0$</p> <p>Сгруппируем слагаемые и применим формулу суммы косинусов $\cos{\alpha} + \cos{\beta} = 2\cos{\frac{\alpha+\beta}{2}}\cos{\frac{\alpha-\beta}{2}}$:</p> <p>$(\cos{10x} + \cos{4x}) + (\cos{8x} + \cos{6x}) = 0$</p> <p>$2\cos{\frac{10x+4x}{2}}\cos{\frac{10x-4x}{2}} + 2\cos{\frac{8x+6x}{2}}\cos{\frac{8x-6x}{2}} = 0$</p> <p>$2\cos{7x}\cos{3x} + 2\cos{7x}\cos{x} = 0$</p> <p>$2\cos{7x}(\cos{3x} + \cos{x}) = 0$</p> <p>Снова применяем формулу суммы косинусов:</p> <p>$2\cos{7x}(2\cos{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}) = 0$</p> <p>$4\cos{7x}\cos{2x}\cos{x} = 0$</p> <p>Получаем три уравнения:</p> <p>1) $\cos{x} = 0 \implies x = \frac{\pi}{2} + \pi k, \quad k \in Z$</p> <p>2) $\cos{2x} = 0 \implies 2x = \frac{\pi}{2} + \pi n \implies x = \frac{\pi}{4} + \frac{\pi n}{2}, \quad n \in Z$</p> <p>3) $\cos{7x} = 0 \implies 7x = \frac{\pi}{2} + \pi m \implies x = \frac{\pi}{14} + \frac{\pi m}{7}, \quad m \in Z$</p> <p>Решения $x = \frac{\pi}{2} + \pi k$ являются подмножеством решений $x = \frac{\pi}{14} + \frac{\pi m}{7}$ (при $m = 3+7k$), поэтому первые можно не указывать в ответе.</p> <p><strong>Ответ:</strong> $x = \frac{\pi}{4} + \frac{\pi n}{2}, \quad n \in Z; \quad x = \frac{\pi}{14} + \frac{\pi m}{7}, \quad m \in Z$.</p> <p><strong>в)</strong></p> <p>Найдем производную функции $f(x) = -\frac{\cos{8x}}{8} - \frac{\cos{2x}}{2}$.</p> <p>$f'(x) = \left(-\frac{1}{8}\cos{8x}\right)' - \left(\frac{1}{2}\cos{2x}\right)' = -\frac{1}{8}(-\sin{8x} \cdot 8) - \frac{1}{2}(-\sin{2x} \cdot 2) = \sin{8x} + \sin{2x}$.</p> <p>Преобразуем функцию $g(x)$ по формуле произведения синуса на косинус $2\sin{\alpha}\cos{\beta} = \sin{(\alpha+\beta)} + \sin{(\alpha-\beta)}$:</p> <p>$g(x) = 2\sin{3x}\cos{x} = \sin(3x+x) + \sin(3x-x) = \sin{4x} + \sin{2x}$.</p> <p>Приравняем $g(x)$ к $f'(x)$:</p> <p>$\sin{4x} + \sin{2x} = \sin{8x} + \sin{2x}$</p> <p>$\sin{4x} = \sin{8x}$</p> <p>$\sin{8x} - \sin{4x} = 0$</p> <p>Используем формулу разности синусов $\sin{\alpha} - \sin{\beta} = 2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}$:</p> <p>$2\cos{\frac{8x+4x}{2}}\sin{\frac{8x-4x}{2}} = 0$</p> <p>$2\cos{6x}\sin{2x} = 0$</p> <p>Получаем два уравнения:</p> <p>1) $\sin{2x} = 0$</p> <p>$2x = \pi n \implies x = \frac{\pi n}{2}, \quad n \in Z$</p> <p>2) $\cos{6x} = 0$</p> <p>$6x = \frac{\pi}{2} + \pi k \implies x = \frac{\pi}{12} + \frac{\pi k}{6}, \quad k \in Z$</p> <p><strong>Ответ:</strong> $x = \frac{\pi n}{2}, \quad n \in Z; \quad x = \frac{\pi}{12} + \frac{\pi k}{6}, \quad k \in Z$.</p> <p><strong>г)</strong></p> <p>Найдем производную функции $f(x) = \frac{\sin{4x}}{8} + \frac{3}{4}x + \frac{\sin{2\pi}}{21}$.</p> <p>Слагаемое $\frac{\sin{2\pi}}{21}$ является константой (равно 0), его производная равна нулю.</p> <p>$f'(x) = \left(\frac{1}{8}\sin{4x}\right)' + \left(\frac{3}{4}x\right)' + 0 = \frac{1}{8}(\cos{4x} \cdot 4) + \frac{3}{4} = \frac{1}{2}\cos{4x} + \frac{3}{4}$.</p> <p>Теперь преобразуем функцию $g(x) = \sin^4{x} + \cos^4{x}$:</p> <p>$g(x) = (\sin^2{x} + \cos^2{x})^2 - 2\sin^2{x}\cos^2{x} = 1^2 - 2(\sin{x}\cos{x})^2 = 1 - 2\left(\frac{\sin{2x}}{2}\right)^2 = 1 - 2\frac{\sin^2{2x}}{4} = 1 - \frac{1}{2}\sin^2{2x}$.</p> <p>Используем формулу понижения степени $\sin^2{\alpha} = \frac{1-\cos{2\alpha}}{2}$:</p> <p>$g(x) = 1 - \frac{1}{2}\left(\frac{1-\cos{4x}}{2}\right) = 1 - \frac{1-\cos{4x}}{4} = \frac{4 - (1-\cos{4x})}{4} = \frac{3+\cos{4x}}{4} = \frac{3}{4} + \frac{1}{4}\cos{4x}$.</p> <p>Приравняем $g(x)$ к $f'(x)$:</p> <p>$\frac{3}{4} + \frac{1}{4}\cos{4x} = \frac{1}{2}\cos{4x} + \frac{3}{4}$</p> <p>$\frac{1}{4}\cos{4x} = \frac{1}{2}\cos{4x}$</p> <p>$\frac{1}{2}\cos{4x} - \frac{1}{4}\cos{4x} = 0$</p> <p>$\frac{1}{4}\cos{4x} = 0$</p> <p>$\cos{4x} = 0$</p> <p>$4x = \frac{\pi}{2} + \pi k, \quad k \in Z$</p> <p>$x = \frac{\pi}{8} + \frac{\pi k}{4}, \quad k \in Z$</p> <p><strong>Ответ:</strong> $x = \frac{\pi}{8} + \frac{\pi k}{4}, \quad k \in Z$.</p>"
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            "field_author" => Illuminate\Database\Eloquent\Collection {#1407
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                0 => App\Models\Term {#1408
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                }
                1 => App\Models\Term {#1409
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                }
                2 => App\Models\Term {#1410
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            "content" => Illuminate\Database\Eloquent\Collection {#1516
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                }
              ]
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            }
            "next" => array:2 [
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                }
                1 => App\Models\Task {#1921
                  #connection: "mysql"
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                }
                2 => App\Models\Task {#2325
                  #connection: "mysql"
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                }
                3 => App\Models\Task {#2729
                  #connection: "mysql"
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                  #primaryKey: "id"
                  #keyType: "int"
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                }
                4 => App\Models\Task {#3133
                  #connection: "mysql"
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                }
                5 => App\Models\Task {#3537
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                }
              ]
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№28 (с. 48)
Условие. №28 (с. 48)
скриншот условия
Алгебра, 10 класс Учебник, авторы: Пак Олег Владимирович, Ардакулы Дархан, Ескендирова Елена Викторовна, издательство Алматыкітап баспасы, Алматы, 2019, Часть 2, страница 48, номер 28, Условие

28. При каких значениях переменной значение функции $g(x)$ равно значению производной функции $f(x)$, если:

а) $f(x) = -\frac{2\cos5x}{5} + \cos\frac{\pi}{3}$, $g(x) = \sin3x + \sin7x$;

б) $f(x) = 2x + \cos^2\frac{4\pi}{17}$, $g(x) = \sin^2 2x + \sin^2 3x + \sin^2 4x + \sin^2 5x$;

в) $f(x) = \frac{\cos8x}{8} - \frac{\cos2x}{2}$, $g(x) = 2\sin3x \cos x$;

г) $f(x) = \frac{\sin4x}{8} + \frac{3}{4}x + \frac{\sin2\pi}{21}$, $g(x) = \sin^4 x + \cos^4 x?$

Решение 2 (rus). №28 (с. 48)

а)

Сначала найдем производную функции $f(x) = -\frac{2\cos{5x}}{5} + \cos{\frac{\pi}{3}}$.

Поскольку $\cos{\frac{\pi}{3}}$ является константой, ее производная равна нулю. Производная от $\cos{5x}$ равна $-5\sin{5x}$.

$f'(x) = \left(-\frac{2}{5}\cos{5x} + \cos{\frac{\pi}{3}}\right)' = -\frac{2}{5}(-\sin{5x} \cdot 5) + 0 = 2\sin{5x}$.

Теперь приравняем значение функции $g(x)$ к значению производной $f'(x)$:

$g(x) = f'(x)$

$\sin{3x} + \sin{7x} = 2\sin{5x}$

Используем формулу суммы синусов: $\sin{\alpha} + \sin{\beta} = 2\sin{\frac{\alpha+\beta}{2}}\cos{\frac{\alpha-\beta}{2}}$.

$2\sin{\frac{3x+7x}{2}}\cos{\frac{7x-3x}{2}} = 2\sin{5x}$

$2\sin{5x}\cos{2x} = 2\sin{5x}$

$2\sin{5x}\cos{2x} - 2\sin{5x} = 0$

$2\sin{5x}(\cos{2x} - 1) = 0$

Это уравнение распадается на два:

1) $\sin{5x} = 0$

$5x = \pi k, \quad k \in Z$

$x = \frac{\pi k}{5}, \quad k \in Z$

2) $\cos{2x} - 1 = 0 \implies \cos{2x} = 1$

$2x = 2\pi n, \quad n \in Z$

$x = \pi n, \quad n \in Z$

Заметим, что множество решений $x = \pi n$ является подмножеством множества решений $x = \frac{\pi k}{5}$ (при $k=5n$). Следовательно, общее решение - это $x = \frac{\pi k}{5}$.

Ответ: $x = \frac{\pi k}{5}, \quad k \in Z$.

б)

Найдем производную функции $f(x) = 2x + \cos^2{\frac{4\pi}{17}}$.

Слагаемое $\cos^2{\frac{4\pi}{17}}$ является константой, поэтому его производная равна нулю.

$f'(x) = (2x)' + (\cos^2{\frac{4\pi}{17}})' = 2 + 0 = 2$.

Приравниваем $g(x)$ к $f'(x)$:

$\sin^2{2x} + \sin^2{3x} + \sin^2{4x} + \sin^2{5x} = 2$

Используем формулу понижения степени $\sin^2{\alpha} = \frac{1-\cos{2\alpha}}{2}$:

$\frac{1-\cos{4x}}{2} + \frac{1-\cos{6x}}{2} + \frac{1-\cos{8x}}{2} + \frac{1-\cos{10x}}{2} = 2$

$1-\cos{4x} + 1-\cos{6x} + 1-\cos{8x} + 1-\cos{10x} = 4$

$4 - (\cos{4x} + \cos{6x} + \cos{8x} + \cos{10x}) = 4$

$\cos{4x} + \cos{6x} + \cos{8x} + \cos{10x} = 0$

Сгруппируем слагаемые и применим формулу суммы косинусов $\cos{\alpha} + \cos{\beta} = 2\cos{\frac{\alpha+\beta}{2}}\cos{\frac{\alpha-\beta}{2}}$:

$(\cos{10x} + \cos{4x}) + (\cos{8x} + \cos{6x}) = 0$

$2\cos{\frac{10x+4x}{2}}\cos{\frac{10x-4x}{2}} + 2\cos{\frac{8x+6x}{2}}\cos{\frac{8x-6x}{2}} = 0$

$2\cos{7x}\cos{3x} + 2\cos{7x}\cos{x} = 0$

$2\cos{7x}(\cos{3x} + \cos{x}) = 0$

Снова применяем формулу суммы косинусов:

$2\cos{7x}(2\cos{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}) = 0$

$4\cos{7x}\cos{2x}\cos{x} = 0$

Получаем три уравнения:

1) $\cos{x} = 0 \implies x = \frac{\pi}{2} + \pi k, \quad k \in Z$

2) $\cos{2x} = 0 \implies 2x = \frac{\pi}{2} + \pi n \implies x = \frac{\pi}{4} + \frac{\pi n}{2}, \quad n \in Z$

3) $\cos{7x} = 0 \implies 7x = \frac{\pi}{2} + \pi m \implies x = \frac{\pi}{14} + \frac{\pi m}{7}, \quad m \in Z$

Решения $x = \frac{\pi}{2} + \pi k$ являются подмножеством решений $x = \frac{\pi}{14} + \frac{\pi m}{7}$ (при $m = 3+7k$), поэтому первые можно не указывать в ответе.

Ответ: $x = \frac{\pi}{4} + \frac{\pi n}{2}, \quad n \in Z; \quad x = \frac{\pi}{14} + \frac{\pi m}{7}, \quad m \in Z$.

в)

Найдем производную функции $f(x) = -\frac{\cos{8x}}{8} - \frac{\cos{2x}}{2}$.

$f'(x) = \left(-\frac{1}{8}\cos{8x}\right)' - \left(\frac{1}{2}\cos{2x}\right)' = -\frac{1}{8}(-\sin{8x} \cdot 8) - \frac{1}{2}(-\sin{2x} \cdot 2) = \sin{8x} + \sin{2x}$.

Преобразуем функцию $g(x)$ по формуле произведения синуса на косинус $2\sin{\alpha}\cos{\beta} = \sin{(\alpha+\beta)} + \sin{(\alpha-\beta)}$:

$g(x) = 2\sin{3x}\cos{x} = \sin(3x+x) + \sin(3x-x) = \sin{4x} + \sin{2x}$.

Приравняем $g(x)$ к $f'(x)$:

$\sin{4x} + \sin{2x} = \sin{8x} + \sin{2x}$

$\sin{4x} = \sin{8x}$

$\sin{8x} - \sin{4x} = 0$

Используем формулу разности синусов $\sin{\alpha} - \sin{\beta} = 2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}$:

$2\cos{\frac{8x+4x}{2}}\sin{\frac{8x-4x}{2}} = 0$

$2\cos{6x}\sin{2x} = 0$

Получаем два уравнения:

1) $\sin{2x} = 0$

$2x = \pi n \implies x = \frac{\pi n}{2}, \quad n \in Z$

2) $\cos{6x} = 0$

$6x = \frac{\pi}{2} + \pi k \implies x = \frac{\pi}{12} + \frac{\pi k}{6}, \quad k \in Z$

Ответ: $x = \frac{\pi n}{2}, \quad n \in Z; \quad x = \frac{\pi}{12} + \frac{\pi k}{6}, \quad k \in Z$.

г)

Найдем производную функции $f(x) = \frac{\sin{4x}}{8} + \frac{3}{4}x + \frac{\sin{2\pi}}{21}$.

Слагаемое $\frac{\sin{2\pi}}{21}$ является константой (равно 0), его производная равна нулю.

$f'(x) = \left(\frac{1}{8}\sin{4x}\right)' + \left(\frac{3}{4}x\right)' + 0 = \frac{1}{8}(\cos{4x} \cdot 4) + \frac{3}{4} = \frac{1}{2}\cos{4x} + \frac{3}{4}$.

Теперь преобразуем функцию $g(x) = \sin^4{x} + \cos^4{x}$:

$g(x) = (\sin^2{x} + \cos^2{x})^2 - 2\sin^2{x}\cos^2{x} = 1^2 - 2(\sin{x}\cos{x})^2 = 1 - 2\left(\frac{\sin{2x}}{2}\right)^2 = 1 - 2\frac{\sin^2{2x}}{4} = 1 - \frac{1}{2}\sin^2{2x}$.

Используем формулу понижения степени $\sin^2{\alpha} = \frac{1-\cos{2\alpha}}{2}$:

$g(x) = 1 - \frac{1}{2}\left(\frac{1-\cos{4x}}{2}\right) = 1 - \frac{1-\cos{4x}}{4} = \frac{4 - (1-\cos{4x})}{4} = \frac{3+\cos{4x}}{4} = \frac{3}{4} + \frac{1}{4}\cos{4x}$.

Приравняем $g(x)$ к $f'(x)$:

$\frac{3}{4} + \frac{1}{4}\cos{4x} = \frac{1}{2}\cos{4x} + \frac{3}{4}$

$\frac{1}{4}\cos{4x} = \frac{1}{2}\cos{4x}$

$\frac{1}{2}\cos{4x} - \frac{1}{4}\cos{4x} = 0$

$\frac{1}{4}\cos{4x} = 0$

$\cos{4x} = 0$

$4x = \frac{\pi}{2} + \pi k, \quad k \in Z$

$x = \frac{\pi}{8} + \frac{\pi k}{4}, \quad k \in Z$

Ответ: $x = \frac{\pi}{8} + \frac{\pi k}{4}, \quad k \in Z$.

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