đặt [tex]sin^2(x+\frac{\pi}{3})=t=>2t^2-5t+2=0<=><=>t=2[/tex] (loại) hoặc [tex]t=\frac{1}{2}[/tex]
ta có: [tex]sin(x+\frac{\pi}{3})=\frac{1}{2}<=>\begin{bmatrix} x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\ x+\frac{\pi}{3}=\pi-\frac{\pi}{6}+k2\pi \end{bmatrix} <=>\begin{bmatrix} x=-\frac{\pi}{3}+k2\pi\\ x=\frac{\pi}{2}+k2\pi \end{bmatrix}[/tex]