Question

If the sum of all solutions of the equation $\sin x+2\cos x=1+\sqrt{3}\cos x$ in $[0,2\pi ]$ is $\frac{k\pi }{6}$ then $\frac{k}{2}$ is `

Answer 1

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

$\Rightarrow \sin (x+\frac{\pi }{12})=\sin \frac{5\pi }{12}$

$\Rightarrow x+\frac{\pi }{12}=\frac{5\pi }{12},x+\frac{\pi }{12}=\pi -\frac{5\pi }{12}$

$x=\frac{\pi }{3}\text{or}\frac{\pi }{2},\text{sum}=\frac{\pi }{3}+\frac{\pi }{2}=\frac{5\pi }{6}\Rightarrow k=5$