Đáp án: $x\in\{k2\pi,\dfrac23\pi+k2\pi\}$
Giải thích các bước giải:
Ta có:
$\sqrt{3}\sin x+\cos x+2\cos (x-\dfrac{\pi}{3})=2$
$\to\dfrac{\sqrt{3}}{2}\sin x+\dfrac12\cos x+\cos (x-\dfrac{\pi}{3})=1$
$\to\sin \left(\dfrac{\pi }{3}\right)\sin x+\cos \left(\dfrac{\pi }{3}\right)\cos x+\cos (x-\dfrac{\pi}{3})=1$
$\to \cos \left(x\right)\cos \left(\dfrac{\pi }{3}\right)+\sin \left(x\right)\sin \left(\dfrac{\pi }{3}\right)+\cos (x-\dfrac{\pi}{3})=1$
$\to \cos \left(x-\dfrac{\pi }{3}\right)+\cos \left(x-\dfrac{\pi }{3}\right)=1$
$\to 2\cos \left(x-\dfrac{\pi }{3}\right)=1$
$\to \cos \left(x-\dfrac{\pi }{3}\right)=\dfrac12$
$\to x-\dfrac{\pi }{3}=\dfrac{\pi}{3}+k2\pi\to x=\dfrac23\pi+k2\pi$
Hoặc $x-\dfrac{\pi}{3}=\dfrac{5\pi}{3}+k2\pi\to x=2\pi+k2\pi=k2\pi$
