Решить уравнение2sinx-2sin^2 x=cos^2 x

[latex]2sinx-2sin ^{2} x=cos ^{2} x \ 2sinx-2sin ^{2} x-cos ^{2}x=0 \ 2sinx-2sin ^{2} x- (1-sin ^{2} x) =0 \ 2 sinx-2sin ^{2} x- 1+sin ^{2} x=0 \ 2sinx-sin ^{2} x-1=0 \ sinx=t \ 2t-t ^{2} -1=0 \ D=8 \ sqrt{D} =2 sqrt{2} \ t_{1}= 1+ sqrt{2} \ t_{2} =1- sqrt{2} \ cosx eq 1+ sqrt{2} \ cosx=1- sqrt{2} \ x=+-(arccos(1- sqrt{2} ))+2 pi n[/latex]