Сos2x+корень из двух* cos(п/2+x)+1=0

Решите задачу:

$cos2x+\sqrt2cos(\frac{\pi}2+x)+1=0\\(cos2x=1-2sin^2x,cos(\frac{\pi}2+x)=-sinx\\1-2sin^2x-\sqrt{2}sinx+1=0\\-2sin^2x-\sqrt{2}sinx+2=0\\2sin^2x+\sqrt2sinx-2=0\\t=sinx,t\in[-1;1]\\2t^2+\sqrt2t-2=0\\D=2+4*2*2=2+16=18\\\\t_1=\frac{-\sqrt{2}-\sqrt{18}}4=\frac{-\sqrt{2}(1+\sqrt{9})}4=-\frac{\sqrt2}4(1+3)=-\frac{\sqrt2}4*4=-\sqrt2,\notin[-1;1]\\\\t_2=\frac{-\sqrt{2}-\sqrt{18}}4=\frac{-\sqrt{2}(1-\sqrt{9})}4=-\frac{\sqrt2}4(1-3)=\frac{\sqrt2}2$

$sinx=\frac{\sqrt2}2\\\\ \left[\begin{array}{ccc}x=\frac{\pi}4+2\pi n;n\in Z\\x=\frac{3\pi}4+2\pi n;n\in Z\end{array}\right$