Đáp án:
a)
$\left\{\begin{array}{l} x=\dfrac{\pi}{18}+k\dfrac{2\pi}9 \\x=-\dfrac{\pi}{30}+k\dfrac{2\pi}5 \end{array} \right .$ $(k\in\mathbb Z)$
Câu b đề thiếu
Lời giải:
a)
$\sin 7x+\sqrt3\cos 7x=\sqrt3\sin 2x+\cos 2x$
$\Rightarrow \dfrac{1}{2}\sin 7x+\dfrac{\sqrt3}{2}\cos 7x=\dfrac{\sqrt3}{2}\sin 2x+\dfrac{1}{2}\cos 2x$
$\Rightarrow\sin(7x+\dfrac{\pi}{3})=\cos(2x-\dfrac{\pi}{3})$
$\Rightarrow \sin(7x+\dfrac{\pi}{3})=\sin(\dfrac{\pi}{2}-2x+\dfrac{\pi}{3})$
$\Rightarrow \left[\begin{array}{l} 7x+\dfrac{\pi}{3}=\dfrac{\pi}{2}-2x+\dfrac{\pi}{3}+k2\pi \\ 7x+\dfrac{\pi}{3}=\pi-\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}+k2\pi \end{array} \right .$
$\Rightarrow \left[\begin{array}{l} x=\dfrac{\pi}{18}+k\dfrac{2\pi}9 \\x=-\dfrac{\pi}{30}+k\dfrac{2\pi}5 \end{array} \right .$ $(k\in\mathbb Z)$
Vậy $\left\{\begin{array}{l} x=\dfrac{\pi}{18}+k\dfrac{2\pi}9 \\x=-\dfrac{\pi}{30}+k\dfrac{2\pi}5 \end{array} \right .$ $(k\in\mathbb Z)$
