Решите уравнение 2sinx*sin3x=cos2x, и найдите корни из промежутка (0;П)

а)

$2\sin x\sin 3x=\cos 2x\mid\cdot ~(-1)\medskip\\{-2\sin x\sin 3x}={-\cos 2x}\medskip\\{-2\sin\dfrac{4x-2x}{2}\sin\dfrac{4x+2x}{2}}={-\cos 2x}\medskip\\\cos 4x-\cos 2x=-\cos 2x\medskip\\\cos 4x=0\medskip\\4x=\dfrac{\pi}{2}+\pi m,~m\in\mathbb{Z}\medskip\\x=\dfrac{\pi}{8}+\dfrac{\pi m}{4},~m\in\mathbb{Z}$

б)

$\left(0;~\pi\right)\medskip\\0<\dfrac{\pi}{8}+\dfrac{\pi m}{4}<\pi;~m\in\mathbb{Z}\medskip\\0<\dfrac{1}{8}+\dfrac{m}{4}<1;~m\in\mathbb{Z}\medskip\\-0{,}5<m<3{,}5;~m\in\mathbb{Z}\Rightarrow m\in\left\{0;~1;~2;~3\right\}$

$x_1=\dfrac{\pi}{8}\medskip\\x_2=\dfrac{\pi}{8}+\dfrac{\pi}{4}=\dfrac{3\pi}{8}\medskip\\x_3=\dfrac{\pi}{8}+\dfrac{2\pi}{4}=\dfrac{5\pi}{8}\medskip\\x_4=\dfrac{\pi}{8}+\dfrac{3\pi}{4}=\dfrac{7\pi}{8}$

Ответ. а) $x=\dfrac{\pi}{8}+\dfrac{\pi m}{4},~m\in\mathbb{Z}$ ; б) $\left\{\dfrac{\pi}{8};~\dfrac{3\pi}{8};~\dfrac{5\pi}{8};~\dfrac{7\pi}{8}\right\}$