|
|
PT $\Leftrightarrow \sin2x=\frac{\pi}{6}+2m\pi$ hoặc $\sin2x=\frac{5\pi}{6}+2n\pi$ , $m,n\in \mathbb{Z}$
Ta có
$-1<\frac{\pi}{6}+2m\pi<1,-1<\frac{5\pi}{6}+2n\pi<1\Rightarrow m=0,n \text { không tồn tại}$.
$\Rightarrow \sin2x=\frac{\pi}{6}$
$\Leftrightarrow 2x=(-1)^k\arcsin\frac{\pi}{6}+k\pi$
$\boxed{x=\frac{1}{2}\left( (-1)^k\arcsin\frac{\pi}{6}+k\pi\right),k\in \mathbb{Z}}$
|