помогите решить cos2x-sin^2(pi/2-x)=-0 25

[2cos^2(x)-1] - cos^2(x)=-1/4;
2cos^2(x)-cos^2(a)=-1/4+1; cos^2(x)=3/4; cosx=+-V3/2.

1) cos x =V3/2; x=+-pi/6+2pin, n---Z.

2) cos x=-V3/2; x=+-5pi/6+2pin, n---Z.