Đáp án:
$4$ nghiệm
Giải thích các bước giải:
$2\sin\left(2x -\dfrac{\pi}{4}\right) = \sqrt2$
$\Leftrightarrow \sin\left(2x -\dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}$
$\Leftrightarrow \left[\begin{array}{l}2x - \dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\2x - \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= \dfrac{\pi}{4} + k\pi\\x = \dfrac{\pi}{2} + k\pi\end{array}\right. \quad (k\in\Bbb Z)$
Ta có:
$-\pi \leq x \leq \dfrac{\pi}{2}$
$\Leftrightarrow \left[\begin{array}{l}-\pi \leq \dfrac{\pi}{4} + k\pi \leq \dfrac{\pi}{2}\\-\pi \leq \dfrac{\pi}{2} + k\pi\leq \dfrac{\pi}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l} -\dfrac{5}{4} \leq k\leq \dfrac{1}{4}\\-\dfrac{3}{2} \leq k\leq 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}k = -1;0\\k = -1;0\end{array}\right. \quad (k\in\Bbb Z)$
$\Rightarrow \left[\begin{array}{l}x = -\dfrac{3\pi}{4}\\x= \dfrac{\pi}{4}\\x = -\dfrac{\pi}{2}\\x = \dfrac{\pi}{2}\end{array}\right.$
$\Rightarrow$ Có $4$ nghiệm thoả mãn đề bài
