Đáp án:
$x = \left\{\dfrac{15\pi}{24};\dfrac{31\pi}{24};\dfrac{49\pi}{24}\right\}$
Giải thích các bước giải:
$\begin{array}{l}2\sin\left(2x + \dfrac{\pi}{6}\right) - \sqrt2=0\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{6}\right)=\dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{6} = \dfrac{\pi}{4} + k2\pi\\2x + \dfrac{\pi}{6} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x= \dfrac{\pi}{24} + k\pi\\x= \dfrac{7\pi}{24} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ \dfrac{5\pi}{6} < x < \dfrac{13\pi}{6}\\ +)\quad \dfrac{5\pi}{6} < \dfrac{\pi}{24} + k\pi < \dfrac{13\pi}{6}\\ \Leftrightarrow \dfrac{19}{24} < k < \dfrac{17}{8}\\ \Rightarrow k = 1;2\\ \Rightarrow x = \left\{\dfrac{25\pi}{24}; \, \dfrac{49\pi}{24}\right\}\\ +)\quad \dfrac{5\pi}{6} < \dfrac{7\pi}{24} + k\pi < \dfrac{13\pi}{6}\\ \Leftrightarrow \dfrac{13}{24} < k < \dfrac{15}{8}\\ \Rightarrow k = 1\\ \Rightarrow x = \dfrac{31\pi}{24}\\ \text{Vậy phương trình có nghiệm}\quad x = \left\{\dfrac{15\pi}{24};\dfrac{31\pi}{24};\dfrac{49\pi}{24}\right\} \end{array}$
