Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 864

\[1)\sin{2x} = 3\sin x \bullet \cos^{2}x\]

\[2\sin x \bullet \cos x - 3\sin x \bullet \cos^{2}x = 0\]

\[\sin x \bullet \cos x \bullet \left( 2 - 3\cos x \right) = 0\]

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

\[x = \pi n.\]

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

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

\[3)\ 2 - 3\cos x = 0\]

\[3\cos x = 2\]

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

\[x = \pm \arccos\frac{2}{3} + 2\pi n.\]

\[Ответ:\ \ \pi n;\ \frac{\pi}{2} + \pi n;\ \]

\[\text{\ \ \ \ } \pm \arccos\frac{2}{3} + 2\pi n.\]

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

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

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

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

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

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

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

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

\[2x = \pm \arccos\frac{1}{2} + 2\pi n =\]

\[= \pm \frac{\pi}{3} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right) = \pm \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \ \frac{\text{πn}}{2};\ \pm \frac{\pi}{6} + \pi n.\]

\[3)\cos{2x} + \cos^{2}x = 0\]

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

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

\[3\cos^{2}x - 1 = 0\]

\[3\cos^{2}x = 1\]

\[\cos^{2}x = \frac{1}{3}\]

\[\cos x = \pm \frac{1}{\sqrt{3}}.\]

\[1)\cos x = - \frac{1}{\sqrt{3}}\]

\[x = \pm \left( \pi - \arccos\frac{1}{\sqrt{3}} \right) + 2\pi n.\]

\[2)\cos x = \frac{1}{\sqrt{3}}\]

\[x = \pm \arccos\frac{1}{\sqrt{3}} + 2\pi n.\]

\[Ответ:\ \pm \arccos\frac{1}{\sqrt{3}} + \pi n.\]

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

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

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

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

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

\[2)\ 2\sin x - \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[2\ tg\ x - 1 = 0\]

\[tg\ x = \frac{1}{2}\]

\[x = arctg\frac{1}{2} + \pi n.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ arctg\frac{1}{2} + \pi n.\]