Question

Number of common solution(s) of the trigonometric equations $\cos 2x+(1-\sqrt{3})=(2-\sqrt{3})\cos x$ and $\sin 3x=2\sin x$ which satisfy the inequality $\sqrt{3}\tan x-1\ge 0$ in $[0,5\pi ]$ is

Answer 1

$2{\cos }^{2}x-1+1-\sqrt{3}=(2-\sqrt{3})\cos x$

$2\cos x(\cos x-1)+\sqrt{3}(\cos x-1)=0$

$(\cos x-1)(2\cos x+\sqrt{3})=0$

$\cos x=1\text{or}\cos x=-\frac{\sqrt{3}}{2}$

$\sin 3x=2\sin x$

$\Rightarrow 3\sin x-4{\sin }^{3}x=2\sin x$

$\sin x=0\text{or}3-4{\sin }^{2}x=2$

$4{\sin }^{2}x=1$

$\sin x=\pm \frac{1}{2}$

common solution in $[0,5\pi ]$ is $0,2\pi ,4\pi ,\frac{5\pi }{6},\frac{7\pi }{6},\frac{17\pi }{6},\frac{19\pi }{6},\frac{29\pi }{6}$

If $\tan x\ge \frac{1}{\sqrt{3}}\Rightarrow x\in [\frac{\pi }{6}+n\pi ,\frac{\pi }{2}+n\pi ];n\in I$

So, common solution:$x=\frac{7\pi }{6},\frac{19\pi }{6}$