Sin2x+sinx+2cosx=cos2x

Решите задачу:

$sin2x+sinx+2cosx=cos2x\\\\2sinx\cdot cosx-(cos^2x-sin^2x)+sinx+2cosx=0\\\\2sinx\cdot cosx-(1-sin^2x)+sin^2x+sinx+2cosx=0\\\\(2sinx\cdot cosx+2cosx)+(2sin^2x+sinx-1)=0\\\\\, [\, 2sin^2x+sinx-1=0\; ,\; D=1+8=9,\\\\(sinx)_1=\frac{-1-3}{4}=-1,\; \; (sinx)_2=\frac{-1+3}{4}=\frac{1}{2}\; ]\\\\2cosx(sinx+1)+2(sinx+1)(sinx+\frac{1}{2})=0\\\\2(sinx+1)(cosx+sinx+\frac{1}{2})=0\\\\a)\; \; sinx+1=0\; ,\; sinx=-1\; ,\; x=-\frac{\pi}{2}+2\pi n\; ,\; n\in Z\\\\b)\; \; cosx+sinx=-\frac{1}{2}\; |:\sqrt2$

$\frac{1}{\sqrt2} cosx+\frac{1}{\sqrt2}sinx=-\frac{1}{2\sqrt2}\\\\sin\frac{\pi}{4}cosx+cos\frac{\pi}{4}sinx=-\frac{1}{2\sqrt2}\\\\sin(x+\frac{\pi}{4})=-\frac{1}{2\sqrt2}\\\\x+\frac{\pi}{4}=(-1)^{k} arcsin(-\frac{1}{2\sqrt2})+\pi k=(-1)^{k+1}arcsin\frac{1}{2\sqrt2}+\pi k,\\\\x=(-1)^{k+1}arcsin\frac{1}{2\sqrt2}-\frac{\pi}{4}+\pi k\; ,\; k\in Z$