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 `

• A. 9.0
• B. 9

Answer 1

Answer: (A), (B)

Given,
$\sin x+2\cos x=1+\sqrt{3}\cos x$
$\Rightarrow (\sin x+2\cos x{)}^2=(1+\sqrt{3}\cos x{)}^2$

$\Rightarrow {\sin }^{2}x+4{\cos }^{2}x+4\sin x\cos x=1+3{\cos }^{2}x+2\sqrt{3}\cos x$

$\Rightarrow 1-{\cos }^{2}x+4{\cos }^{2}x+4\sin x\cos x=1+3{\cos }^{2}x+2\sqrt{3}\cos x$

$\Rightarrow 3{\cos }^{2}x+4\sin x\cos x=3{\cos }^{2}x+2\sqrt{3}\cos x)$
$\Rightarrow \cos x(2-\sqrt{3}\sin x)=0$
$\Rightarrow \cos x=0\text{or}(2-\sqrt{3}\sin x)=0$
$\Rightarrow \cos x=0\text{or}\sin x=\frac{\sqrt{3}}{2}$
$\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}\text{or}\frac{\pi }{3},\frac{2\pi }{3}$
Sum of solutions is
$\frac{\pi }{2}+\frac{3\pi }{2}+\frac{\pi }{3}+\frac{2\pi }{3}=\frac{4\pi }{2}+\frac{3\pi }{3}=3\pi$
Given,
$\frac{k\pi }{6}=3\pi$
$\Rightarrow k=18$
$\Rightarrow \frac{k}{2}=9$