Помогите решить уравнение cos 2x = sin (7pi/2 - x)?

$cosx^{2} - sinx^{2} -cosx=0$ ,
$cosx^{2}-(1- cosx^{2})-cosx=0$ ,
$cos x^{2} -1+cos x^{2} -cosx=0$ ,
$2cos x^{2}-cosx-1=0$ ,
$cos x=t$ ,
$2 t^{2} -t-1=0$ ,
D=1+8=9 ,
$t= \frac{1+3}{4}$ ,
$t= \frac{1-3}{4}$ ,
$t_{1} = 1$ ,
$t_{2} = -\frac{1}{2}$
cos=1 , cos=-1/2
$x_{1} = \pi n , x_{2} = \frac{2 \pi }{3} +2 \pi n , x_{3} = - \frac{2 \pi }{3} +2 \pi n .$