Lời giải:
\(\sin ^2(\frac{\pi}{6}-x)=\frac{1}{4}\)
\(\Rightarrow \left[\begin{matrix} \sin (\frac{\pi}{6}-x)=\frac{1}{2}\\ \sin (\frac{\pi}{6}-x)=\frac{-1}{2}\end{matrix}\right.\)
Nếu \(\sin (\frac{\pi}{6}-x)=\frac{1}{2}\Rightarrow \left[\begin{matrix} \frac{\pi}{6}-x=\frac{\pi}{6}-2k\pi \\ \frac{\pi}{6}-x=\frac{5\pi}{6}-2k\pi \end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=2k\pi \\ x=2k\pi-\frac{2}{3}\pi \end{matrix}\right.\) với $k$ nguyên.
Nếu \(\sin (\frac{\pi}{6}-x)=\frac{-1}{2}\Rightarrow \left[\begin{matrix} \frac{\pi}{6}-x=\frac{-\pi}{6}-2k\pi \\ \frac{\pi}{6}-x=\frac{7\pi}{6}-2k\pi \end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{\pi}{3}+2k\pi \\ x=(2k-1)\pi\end{matrix}\right.\) với $k$ nguyên.
Gộp cả 2TH trên lại ta suy ra \(x=n\pi \) hoặc \(x=n\pi+\frac{\pi}{3}\) với $n$ là số nguyên bất kỳ.
