4cos^2x = 8sin(П/2-х)+5 помогите решить!

Решите задачу:

$4cos^2x=8sin(\frac{\pi}{2}-x)+5\\\\.\quad sin(\frac{\pi}{2}-x)=cosx\\\\4cos^,,2x-8cosx-5=0\\\\t=cosx\; ;\; \; -1 \leq t \leq 1\\\\4t^2-8t-5=0\\\\D=64+80=144\; ;\; \; t_1=\frac{5}{2}\ \textgreater \ 1\; ;\; \; t_2=-\frac{1}{2}\\\\cosx=-\frac{1}{2}\; ;\; \; x=\pm arccos(-\frac{1}{2})+2\pi n,n\in Z\\\\x=\pm (\pi -\frac{\pi}{3})+2\pi n=\pm \frac{2\pi}{3}+2\pi n,\; n\in Z$