JEE Main 2025 Apr 3 Shift 2, Mathematics Q16: Trigonometric Equations

Rationalise the right-hand side: $\frac{-4}{1+\sqrt{3}}\cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{-4(\sqrt{3}-1)}{2}=-2(\sqrt{3}-1)=2(1-\sqrt{3})$.

Substitute ${\cos }^{2}x=1-{\sin }^{2}x$:

$(4-\sqrt{3})\sin x-2\sqrt{3}(1-{\sin }^{2}x)=2(1-\sqrt{3})$.

$2\sqrt{3}{\sin }^{2}x+(4-\sqrt{3})\sin x-2\sqrt{3}-2+2\sqrt{3}=0$.

$2\sqrt{3}{\sin }^{2}x+(4-\sqrt{3})\sin x-2=0$.

Let $s=\sin x$. Factor: $(2s-1)(\sqrt{3}s+2)=0$. The roots are $s=1/2$ and $s=-2/\sqrt{3}$. The latter has $|s|>1$, so reject.

Hence $\sin x=1/2$, giving $x=\frac{\pi }{6}+2k\pi$ or $x=\frac{5\pi }{6}+2k\pi$.

List solutions in $[-2\pi ,5\pi /2]$ (length $9\pi /2$): $-\frac{11\pi }{6},-\frac{7\pi }{6},\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6}$. That is 5 solutions.