Đáp án:
$\left[\begin{array}{l}x= \dfrac{3\pi}{4} + k6\pi\\x=\dfrac{9\pi}{4} + k6\pi \end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$\sin^2\dfrac{x}{3} + \dfrac{\sqrt2}{2}\sin\dfrac{x}{3} - 1 = 0$
$\Leftrightarrow \left(\sin\dfrac{x}{3} - \dfrac{\sqrt2}{2}\right)\cdot\left(\sin\dfrac{x}{3} +\sqrt2\right) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin\dfrac{x}{3} = \dfrac{\sqrt2}{2}\quad (nhận)\\\sin\dfrac{x}{3} =-\sqrt2 \qquad (loại)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{x}{3} = \dfrac{\pi}{4} + k2\pi\\\dfrac{x}{3} =\dfrac{3\pi}{4} + k2\pi \end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= \dfrac{3\pi}{4} + k6\pi\\x=\dfrac{9\pi}{4} + k6\pi \end{array}\right.\quad (k\in \Bbb Z)$
