pt <=> $\frac{1+\cos (\frac{2\pi }{3}+2x)}{2}$ + $\frac{1+\cos (\frac{2\pi }{3}-2x)}{2}$ = $\frac{4+\sin x}{2}$ <=> $\cos (\frac{2\pi }{3}+2x)$ + $\cos (\frac{2\pi }{3}-2x)$ = 2 + $\sin x$
<=> 2.$\cos \frac{2\pi }{3}$.$\cos 2x$ = 2 + $\sin x$
<=> - $\cos 2x$ = 2 + $\sin x$
<=> 2$\sin ^{2}x$ - $\sin x$ - 3 = 0
<=> $\left[ \sin x{=\frac{3}{2}(loại}) hoặc \right.\sin x=-1$
$\sin x$ = -1 <=> x = $\frac{-\pi }{2}$ + k2$\pi $