JEE Main 2022 (Online) 24th June Evening ShiftQuestion 1
The number of solutions of the equation $\cos\left(x+\frac{\pi}{3}\right)\cos\left(\frac{\pi}{3}-x\right)=\frac{1}{4}\cos^{2}2x$, $x\in[-3\pi,3\pi]$ is:
Trigonometric Ratio & Identities
Solution
Concept. The left side becomes $\frac12\cos2x-\frac14$; the equation is a perfect square in $\cos2x$. Formulas used. (1) $2\cos A\cos B=\cos(A+B)+\cos(A-B)$Step 1 - Left side. By (1), $$\cos\left(x+\tfrac\pi3\right)\cos\left(\tfrac\pi3-x\right)=\tfrac12\left[\cos\tfrac{2\pi}{3}+\cos2x\right]=\tfrac12\cos2x-\tfrac14.$$ Step 2 - Equation. Multiply by $4$: $2\cos2x-1=\cos^22x$, i.e. $(\cos2x-1)^2=0$. So $\cos2x=1$ and $x=k\pi$. Step 3 - Count. In $[-3\pi,3\pi]$: $k=-3,\dots,3$, i.e. $7$. Answer: (D)