\(sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=\dfrac{-1}{2}.\\ TXD:D=R.\\ \Leftrightarrow sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=sin\dfrac{-\pi}{6}.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{\pi}{4}=\dfrac{-\pi}{6}+k2\pi.\\\dfrac{x}{2}-\dfrac{\pi}{4}=\pi-\dfrac{-\pi}{6}+k2\pi.\end{matrix}\right.\) \(\left(k\in Z\right).\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-3\pi=-2\pi+k24\pi.\\6x-3\pi=12\pi-2\pi+k24\pi.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k4\pi.\\x=\dfrac{13}{6}\pi+k4\pi.\end{matrix}\right.\)
Lời giải:
$\sin (\frac{x}{2}-\frac{\pi}{4})=\frac{-1}{2}=\sin (\frac{-\pi}{6})$
$\Rightarrow \frac{x}{2}-\frac{\pi}{4}=\frac{-\pi}{6}+2k\pi$ hoặc $\frac{x}{2}-\frac{\pi}{4}=\pi +\frac{\pi}{6}+2k\pi$ với $k$ nguyên
$\Rightarrow x=\frac{\pi}{12}+4k\pi$ hoặc $x=\frac{17\pi}{6}+4k\pi$ với $k$ nguyên bất kỳ.
