Đáp án:
$\left[\begin{array}{l} x= \dfrac{7\pi}{12}+ k 2 \pi(k \in \mathbb{Z})\\x= \dfrac{11\pi}{12}+ k 2 \pi(k \in \mathbb{Z})\end{array} \right..$
Giải thích các bước giải:
$\sin x-\cos x=\dfrac{\sqrt{6}}{2}\\ \Leftrightarrow \sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\\ \Leftrightarrow\sin \left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\\ \Leftrightarrow \sin \left(x-\dfrac{\pi}{4}\right)=\sin \dfrac{\pi}{3}\\ \Leftrightarrow \left[\begin{array}{l} x-\dfrac{\pi}{4}= \dfrac{\pi}{3}+ k 2 \pi(k \in \mathbb{Z})\\x-\dfrac{\pi}{4}= \dfrac{2\pi}{3}+ k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x= \dfrac{7\pi}{12}+ k 2 \pi(k \in \mathbb{Z})\\x= \dfrac{11\pi}{12}+ k 2 \pi(k \in \mathbb{Z})\end{array} \right..$
