Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1376

\[\boxed{\mathbf{1376}\mathbf{.}}\]

\[1)\sin^{4}x - \cos^{4}x + 2\cos^{2}x = \cos{2x}\]

\[2\sin^{2}x = 0\]

\[\sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[Ответ:\ \ \pi n.\]

\[2)\ 2\sin^{2}x - \cos^{4}x = 1 - \sin^{4}x\]

\[2\sin^{2}x - 1 - \left( \cos^{4}x - \sin^{4}x \right) = 0\]

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

\[- 2\left( \cos^{2}x - \sin^{2}x \right) = 0\]

\[- 2\cos{2x} = 0\]

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

\[2x = \arccos 0 + \pi n\]

\[2x = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right)\]

\[x = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]