Đáp án:
\(S =\left\{-\pi;\ -\dfrac{2\pi}{3};\ \dfrac{\pi}{3};\ \dfrac{2\pi}{3};\ \dfrac{5\pi}{3};\ 2\pi\right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \sin\left(\dfrac{3x}{2} - \dfrac{\pi}{4}\right) =\dfrac{\sqrt2}{2}\\
\Leftrightarrow \left[\begin{array}{l}\dfrac{3x}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\\dfrac{3x}{2} - \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{3} + k\dfrac{4\pi}{3}\\x = \dfrac{2\pi}{3} + k\dfrac{4\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Ta có:}\\
\quad -\pi \leqslant x \leqslant 2\pi\\
\Leftrightarrow \left[\begin{array}{l}-\pi \leqslant \dfrac{\pi}{3} + k\dfrac{4\pi}{3} \leqslant 2\pi\\-\pi \leqslant \dfrac{2\pi}{3} + k\dfrac{4\pi}{3} \leqslant 2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}-1 \leqslant k \leqslant \dfrac54\\-\dfrac54 \leqslant k \leqslant 1 \end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}k\in \{-1;0;1\}\\k \in \{-1;0;1\}\end{array}\right.\\
\Rightarrow \left[\begin{array}{l}x = -\pi\\x = \dfrac{\pi}{3}\\x = \dfrac{5\pi}{3}\\x = -\dfrac{2\pi}{3}\\x = \dfrac{2\pi}{3}\\x = 2\pi\end{array}\right.\\
\text{Vậy}\ S =\left\{-\pi;\ -\dfrac{2\pi}{3};\ \dfrac{\pi}{3};\ \dfrac{2\pi}{3};\ \dfrac{5\pi}{3};\ 2\pi\right\}
\end{array}\)
