$cos^{2}\left (x+\frac{r}{3} \right ) +cos^{2}\left ( x+\frac{2r}{3} \right )$ =$\frac{1}{2}$(sinx+1)$
<=>$\frac{cos^{2}x}{4}+\frac{3sin^{2}}{4}+\frac{\sqrt{3}sinxcosx}{2}=\frac{1+sinx}{2}$
<=> $1+2sin^{2}x+\sqrt{3}sin2x=2+2sinx$
<=>$2-cos2x+\sqrt{3}sin2x=2+2sinx$
<=>$sin(2x-\frac{r}{6})=sinx$
kết quả bài này là sinx=0 và sinx=1/2
$\cos^2\left ( x+\frac{\pi}{3} \right )+\cos^2\left ( x+\frac{2\pi}{3} \right )^2=\frac{1}{2}\left ( \sin x+1 \right )$
$\Leftrightarrow \left ( \frac{\cos x}{2}-\frac{\sqrt3\sin x}{2} \right )^2+\left ( -\frac{\cos x}{2}-\frac{\sqrt3\sin x}{2} \right )^2=\frac{1+\sin x}{2}$
$\Leftrightarrow \cos^2x+3\sin^2x=1+\sin x\Leftrightarrow 2\sin^2x=\sin x$
$\Leftrightarrow \sin x=0$ hoặc $\sin x=\frac{1}{2}$
$\Leftrightarrow x=k\pi$ hoặc $x=\frac{\pi}{6}+2k\pi$ hoặc $x=\frac{5\pi}{6}+2k\pi$ ($k\in\mathbb{Z}$)