cos2x=sin(x+pi/2) не уверен в своем решении, хочу свериться с вами

cos2x = (cosx)^2 - (sinx)^2

sin(x+п/2) = cosx

(cosx)^2 - (sinx)^2 - cosx = 0

(cosx)^2 - (1-(cosx)^2) - cosx = 0

2(cosx)^2 - cosx - 1 = 0

cosx = y

2y^2 - y - 1 = 0

D = 1-4*2*(-1) = 9

y1 = (1+3)/4 _____ y1 = 1 _____ cosx1 = 1 _____ x1 = 2пN 

y2 = (1-3)/4 _____ y2 = -1/2 _____ cosx2 = -1/2 _____ x2 = +-2п/3 + 2пN