JEE Main 2021 (Online) 1st September Evening ShiftQuestion 1
If $n$ is the number of solutions of the equation $2\cos x\left(4\sin\left(\frac\pi4+x\right)\sin\left(\frac\pi4-x\right)-1\right)=1$, $x\in[0,\pi]$ and $S$ is the sum of all these solutions, then the ordered pair $(n,S)$ is :
Trigonometric Ratio & Identities
Solution
Concept. The bracket is $2\cos2x-1$; together with $2\cos x$ it becomes $2\cos3x$. Formulas used. (1) $2\sin A\sin B=\cos(A-B)-\cos(A+B)$(2) $\cos3x=4\cos^3x-3\cos x$Step 1 - Bracket. By (1), $4\sin\left(\frac\pi4+x\right)\sin\left(\frac\pi4-x\right)=2\left(\cos2x-\cos\frac\pi2\right)=2\cos2x$. Step 2 - Simplify. $2\cos x(2\cos2x-1)=2\cos x(4\cos^2x-3)=2\cos3x$ by (2). So $\cos3x=\frac12$. Step 3 - Solve. $3x\in[0,3\pi]$: $3x=\frac\pi3,\frac{5\pi}3,\frac{7\pi}3$, i.e. $x=\frac\pi9,\frac{5\pi}9,\frac{7\pi}9$. Step 4 - Value. $n=3$, $S=\frac{13\pi}{9}$. Answer: (B)