$sinx + 2cosx +cos2x - sin2x = 0$
$⇔sinx+2cosx+cos^2x-sin^2x-2sinxcosx=0$
$⇔sinx(1-sinx)+2cosx(1-sinx)+1-sin^2x=0$
$⇔(sinx+2cosx)(1-sinx)+(1-sin^2x)=0$
$⇔(1-sinx)(sinx+2cosx+1+sinx)=0$
$⇔(1-sinx)(2cosx+2sinx)=0$
$⇔$\(\left[ \begin{array}{l}1-sinx=0\\2(cosx+sinx)=-1\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}sinx=1\\cosx+sinx=-\dfrac{1}{2}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=±\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\sin(x+\dfrac{\pi}{4})=-\dfrac{\sqrt[]{2}}{4}(*)\end{array} \right.\)
$(*)⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}-arcsin(\dfrac{\sqrt[]{2}}{4})+k2\pi\\x=\dfrac{5\pi}{4}+arcsin(\dfrac{\sqrt[]{2}}{4})+k2\pi\end{array} \right.\)
