JEE Main 2025 — 3 April, Evening Shift — Question 32

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

Let $\sin x=t \Rightarrow \cos ^{2} x=1-t^{2}$

$(4-\sqrt{3}) t-2 \sqrt{3}\left(1-t^{2}\right)=\frac{-4}{(1+\sqrt{3})}$

$\Rightarrow \quad 2 \sqrt{3} t^{2}+(4-\sqrt{3}) t-2 \sqrt{3}+\frac{4}{1+\sqrt{3}}=0$

$2 \sqrt{3} t^{2}+(4-\sqrt{3}) t+\frac{(-2 \sqrt{3}-2)}{(1+\sqrt{3})}=0$ $2 \sqrt{3} t^{2}+(4-\sqrt{3}) t-2=0$

$\Rightarrow t=\frac{(\sqrt{3}-4) \pm \sqrt{19-8 \sqrt{3}+8(2 \sqrt{3})}}{4 \sqrt{3}}$

$t=\frac{(\sqrt{3}-4) \pm \sqrt{19+8 \sqrt{3}}}{4 \sqrt{3}}=\frac{(\sqrt{3}-4) \pm(\sqrt{3}+4)}{4 \sqrt{3}}$

$=\left(\frac{2 \sqrt{3}}{4 \sqrt{3}}\right)$ or $\frac{-8}{4 \sqrt{3}}$

$\Rightarrow \sin x=\frac{1}{2}$ or $\frac{-2}{\sqrt{3}}<-1 \Rightarrow$ only $\sin x=\frac{1}{2}$

$\Rightarrow 5$ solutions in $x \in\left[-2 \pi, \frac{5 \pi}{2}\right]$