JEE Advanced 2016 · Paper 1, Mathematics question 39

Clear the denominators. Multiply by $\sin x\cos x$ (non-zero on $S$):
$$\sqrt3\sin x+\cos x+2(\sin^2x-\cos^2x)=0\implies\sqrt3\sin x+\cos x=2\cos2x.$$

Compound angle. This is $2\cos\left(x-\frac\pi3\right)=2\cos2x$, so
$$2x=2k\pi\pm\left(x-\frac\pi3\right).$$
This gives $x=2k\pi-\frac\pi3$ or $x=\frac{2k\pi}3+\frac\pi9$.

Solutions in $S$. $-\dfrac\pi3$, $\dfrac\pi9$, $\dfrac{7\pi}9$ and $-\dfrac{5\pi}9$.

Sum. $\dfrac{-3\pi+\pi+7\pi-5\pi}9=0$.

(C)