JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 71

At $x=0$, $\sin y=0$, so $y=0$. Differentiating, $\cos y\,y'=\sin\left(\frac\pi3+y\right)+x\cos\left(\frac\pi3+y\right)y'$, which gives $y'=\frac{\sqrt3}2$ at the origin.

The normal has slope $-\frac2{\sqrt3}$: $2x+\sqrt3y=0$.