JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Trigonometric Equations

Where $\cot x\ge 0$ the left side is $\cot x$, and the equation collapses to $\frac{1}{\sin x}=0$, which no $x$ satisfies. Where $\cot x<0$ the left side is $-\cot x$, so

In $[0,2\pi ]$ this holds at $x=\frac{2\pi }{3}$ and $x=\frac{4\pi }{3}$. At $x=\frac{2\pi }{3}$ the sine is positive and the cosine negative, so $\cot x<0$ as the case demands; at $x=\frac{4\pi }{3}$ both are negative, so $\cot x>0$ and that value belongs to the other case, where no solution exists. The equation has exactly $1$ solution.