Question

2sin^2x-sin2x-2cos2x=0

Answer 1

$2sin^2x-sin2x-2cos2x=0 \\ 2sin^2x-2sinxcosx-2cos^2x+2sin^2x=0 \\ -2sinxcosx-2cos^2x=0 \\ -2(sinxcosx+cos^2x)=0(:-2) \\ sinxcosx+cos^2x=0 \\ cosx(sinx+cosx)=0 \\ \\ cosx=0 \\ x= \frac{ \pi }{2} + \pi k, k \in Z \\ \\ sinx+cosx=0 \\ cosx=-sinx \\ tgx=-1 \\ x=- \frac{ \pi }{4} + \pi k, k\in Z$