ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1373

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

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

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

\[tg\ x - 1 = 0\]

\[tg\ x = 1\]

\[x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n.\]

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

\[2\cos x = - 1\]

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

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

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

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

\[Ответ:\ \ \frac{\pi}{4} + \pi n;\ \ \pm \frac{2\pi}{3} + 2\pi n.\]

\[2)\ 2\cos{2x} = \sqrt{6}\left( \cos x - \sin x \right)\ \]

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

\[1 - tg\ x = 0\]

\[tg\ x = 1\]

\[x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n.\]

\[2\cos x + 2\sin x - \sqrt{6} = 0\]

\[\sin\left( \frac{\pi}{2} - x \right) + \sin x = \frac{\sqrt{6}}{2}\]

\[2 \bullet \sin\frac{\frac{\pi}{2} - x + x}{2} \bullet \cos\frac{\frac{\pi}{2} - x - x}{2} = \frac{\sqrt{6}}{2}\]

\[2 \bullet \sin\frac{\pi}{4} \bullet \cos\left( \frac{\pi}{4} - x \right) = \frac{\sqrt{6}}{2}\]

\[\sqrt{2} \bullet \cos\left( \frac{\pi}{4} - x \right) = \frac{\sqrt{3}}{\sqrt{2}}\]

\[\cos\left( x - \frac{\pi}{4} \right) = \frac{\sqrt{3}}{2}\]

\[x - \frac{\pi}{4} = \pm \arccos\frac{\sqrt{3}}{2} + 2\pi n =\]

\[= \pm \frac{\pi}{6} + 2\pi n;\]

\[x_{1} = - \frac{\pi}{6} + 2\pi n + \frac{\pi}{4} = \frac{\pi}{12} + 2\pi n;\]

\[x_{2} = + \frac{\pi}{6} + 2\pi n + \frac{\pi}{4} = \frac{5\pi}{12} + 2\pi n.\]

\[Ответ:\ \ \frac{\pi}{4} + \pi n;\ \]

\[\frac{\pi}{12} + 2\pi n;\ \ \frac{5\pi}{12} + 2\pi n.\]