QJEEM21Ma7a6b: JEE Main 2021 Mathematics - Class 11 JEEM21Ma7a6b Solution: If n is the number of solutions of the equation and S is the…

Question

Verified Solution • 2025–26 Session moderate

If n is the number of solutions of the equation

2\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - 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 :

Accepted Answer

2\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - x} \right) - 1} \right) = 1

2\cos x\left( {4\left( {{{\sin }^2}{\pi \over 4} - {{\sin }^2}x} \right) - 1} \right) = 1

2\cos x\left( {4\left( {{1 \over 2} - {{\sin }^2}x} \right) - 1} \right) = 1

2\cos x(2 - 4{\sin ^2}x - 1) = 1

2\cos x(1 - 4{\sin ^2}x) = 1

2\cos x(4{\cos ^2}x - 3) = 1

4{\cos ^3}x - 3\cos x = {1 \over 2}

\cos 3x = {1 \over 2}

x \in [0,\pi ]

\therefore

3x \in [0,3\pi ]

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