Đáp án:
9sin²2x - 6sin2x + 1 =0
⇔ (3 sin2x - 1)² =0
⇔3sin2x - 1 =0
⇔sin2x = 1/3
⇔ \(\left[ \begin{array}{l}2x=arcsin\frac{1}{3}+ k2\pi \\2x=\pi -arcsin\frac{1}{3}+ k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=arcsin\frac{1}{6}+ k\pi \\x=\pi/2 -arcsin\frac{1}{6}+ k\pi\end{array} \right.\)