Cos2x=sin(3pi/2-x) решить уравнение

Cos2x = sin(3pi/2 - x)
cos2x = - cosx 
cos2x + cosx =  0
2cos^2x + cosx - 1 = 0 
пусть cosx = t, t ∈ [ -1; 1] 
2t^2 + t - 1 = 0 
D = 1 + 4*2 = 9 
t1 = ( - 1 +3)/4 = 2/4 = 1/2
t2 = ( - 1 - 3)/4 = - 4/4 = - 1

cosx = 1/2
x = ± arccos(1/2) + 2pik
x = ± pi/3 + 2pik.k ∈ Z

cosx = - 1
x = pi + 2pik.k ∈ Z 

ответ:
x = ± pi/3 + 2pik.k ∈ Z
x = pi + 2pik.k ∈ Z