\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=k\pi\)
Do \(-1\le sin\left(x+\frac{\pi}{3}\right)\le1\Rightarrow-1\le k\pi\le1\)
\(\Rightarrow-\frac{1}{\pi}\le k\le\frac{1}{\pi}\Rightarrow k=0\)
\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\frac{\pi}{3}=l\pi\)
\(\Leftrightarrow x=-\frac{\pi}{3}+l\pi\)
Lời giải:
$\sin [\sin (x+\frac{\pi}{3})]=0$
$\Rightarrow \sin (x+\frac{\pi}{3})=k\pi$ với $k\in\mathbb{Z}$
Mà: $\sin (x+\frac{\pi}{3})\in [-1;1]$ nên $k\pi [-1;1]$
Do $k$ nguyên nên $k=0$
$\Rightarrow \sin (x+\frac{\pi}{3})=0$
$\Rightarrow x+\frac{\pi}{3}=m\pi$ với $m$ nguyên
$\Rightarrow x=\pi (m-\frac{1}{3})$ với $m$ nguyên.
