Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 26

\[\boxed{\mathbf{26.}}\]

\[\textbf{а)}\ \frac{\sin{2x}}{\cos{2x}} = \frac{\sin x}{\cos x}\]

\[\frac{\sin{2x} \cdot \cos x - \cos{2x}\sin x}{\cos{2x}\cos x} = 0\]

\[\frac{\sin x}{\cos{2x}\cos x} = 0\]

\[\sin x = 0\]

\[x = \pi n.\]

\[Проверка:\]

\[\cos(2\pi n) \cdot \cos\left( \text{πn} \right) =\]

\[= 1 \cdot ( - 1) = - 1 \neq 0\]

\[x = \pi n - корень.\]

\[Ответ:\ x = \pi n.\]

\[\textbf{б)}\ \frac{\sin{2x}}{\cos{2x}} = \frac{\sin x}{\cos x}\]

\[\frac{\sin{2x} \cdot \cos x + \cos{2x}\sin x}{\cos{2x}\cos x} = 0\]

\[\frac{\sin{3x}}{\cos{2x}\cos x} = 0\]

\[\sin{3x} = 0\]

\[3x = \pi n\]

\[x = \frac{\text{πn}}{3}.\]

\[Проверка:\]

\[\cos\left( \frac{2\pi n}{3} \right) \cdot \cos\left( \frac{\text{πn}}{3} \right) =\]

\[= - \frac{1}{2} \cdot \frac{1}{2} = - \frac{1}{4} \neq 0;\]

\[x = \frac{\text{πn}}{3} - корень.\]

\[Ответ:x = \frac{\text{πn}}{3}.\]

\[\textbf{в)}\ \frac{\sin x}{\cos x} = \frac{\sin{4x}}{\cos{4x}}\]

\[\frac{\sin x\cos{4x} - \sin{4x}\cos x}{\cos x\cos{4x}} = 0\]

\[\frac{\sin{( - 3x)}}{\cos x\cos{4x}} = 0\]

\[- \sin{3x} = 0\]

\[3x = \pi n\]

\[x = \frac{\text{πn}}{3}.\]

\[Проверка:\]

\[\cos\frac{\text{πn}}{3} \cdot \cos{4 \cdot \frac{\text{πn}}{3}} = \frac{1}{2} \cdot \left( - \frac{1}{2} \right) =\]

\[= - \frac{1}{4} \neq 0;\]

\[x = \frac{\text{πn}}{3} - корень.\]

\[Ответ:x = \frac{\text{πn}}{3}.\]

\[\textbf{г)}\ \frac{\sin x}{\cos x} = - \frac{\sin{4x}}{\cos{4x}}\]

\[\frac{\sin x\cos{4x} + \sin{4x}\cos x}{\cos x\cos{4x}} = 0\]

\[\frac{\sin{(5x)}}{\cos x\cos{4x}} = 0\]

\[\sin{5x} = 0\]

\[5x = \pi n\]

\[x = \frac{\text{πn}}{5}.\]

\[Проверка:\]

\[\cos\frac{\text{πn}}{5} \cdot \cos{5 \cdot \frac{\text{πn}}{3}} \neq 0;\]

\[x = \frac{\text{πn}}{5} - корень.\]

\[Ответ:x = \frac{\text{πn}}{5}.\]