Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 512

\[\boxed{\mathbf{512}\mathbf{.}}\]

\[1)\sin{2x} - 2\cos x = 0\]

\[2\sin x\cos x - 2\cos x = 0\]

\[2\cos x\left( \sin x - 1 \right) = 0\]

\[1)\ 2\cos x = 0\]

\[\cos x = 0\]

\[x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[2)\ \sin x - 1 = 0\]

\[\sin x = 1\]

\[x = \arcsin 1 + 2\pi n\]

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

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]

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

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

\[\cos^{2}x = 1\]

\[\cos x = \pm 1\]

\[x_{1} = \pi - \arccos 1 + 2\pi n =\]

\[= \pi + 2\pi n\]

\[x_{2} = \arccos 1 + 2\pi n\]

\[x = 2\pi n\]

\[Ответ:\ \ \pi n.\]

\[3)\ 4\cos x = \sin{2x}\]

\[4\cos x - \sin{2x} = 0\]

\[4\cos x - 2\sin x \bullet \cos x = 0\]

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

\[1)\ 2\cos x = 0\]

\[\cos x = 0\]

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

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

\[2)\ 2 - \sin x = 0\]

\[\sin x = 2\]

\[корней\ нет\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]

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

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

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

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

\[\cos x = 0\]

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

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

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]

\[5)\sin{\frac{x}{2}\cos\frac{x}{2}} + \frac{1}{2} = 0\]

\[\frac{1}{2} \bullet 2\sin\frac{x}{2}\cos\frac{x}{2} + \frac{1}{2} = 0\]

\[\frac{1}{2}\sin\left( 2 \bullet \frac{x}{2} \right) + \frac{1}{2} = 0\]

\[\frac{1}{2}\sin x = - \frac{1}{2}\]

\[\sin x = - 1\]

\[x = - \arcsin 1 + 2\pi n\]

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

\[Ответ:\ - \frac{\pi}{2} + 2\pi n.\]

\[6)\cos^{2}\frac{x}{2} = \sin^{2}\frac{x}{2}\]

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

\[\cos\left( 2 \bullet \frac{x}{2} \right) = 0\]

\[\cos x = 0\]

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

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

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]