JEE Main 2025 (Online) 2nd April Morning ShiftQuestion 1
If $\theta\in[-2\pi,2\pi]$, then the number of solutions of $2\sqrt{2}\cos^{2}\theta+\left(2-\sqrt{6}\right)\cos\theta-\sqrt{3}=0$, is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Solve the quadratic in $c=\cos\theta$, then count in $[-2\pi,2\pi]$. Formulas used. (1) Product of roots of $ac^2+bc+d$ is $\frac da$Step 1 - Roots. $c=\frac{\sqrt3}{2}$ works: $\frac{3\sqrt2}{2}+\sqrt3-\frac{3\sqrt2}{2}-\sqrt3=0$. By (1) the product is $-\frac{\sqrt3}{2\sqrt2}$, so the other root is $-\frac{1}{\sqrt2}$. Step 2 - $\cos\theta=\frac{\sqrt3}2$. $\theta=\pm\frac\pi6,\pm\frac{11\pi}6$: $4$ values. Step 3 - $\cos\theta=-\frac1{\sqrt2}$. $\theta=\pm\frac{3\pi}4,\pm\frac{5\pi}4$: $4$ values. Step 4 - Total. $8$. Answer: (A)