2cosx(4sin(4π+x)sin(4π−x)−1)=1
⇒2cosx(2cos(2x)−2cos(2π)−1)=1
⇒2cosx(2.(2cos2x−1)−1)=1
⇒2cosx(4cos2x−3)=1
⇒4cos3x−3cosx=21
⇒cos3x=21
⇒3x=3π,35π,37π
⇒x=9π,95π,97π
No. of solutions =n=3
Sum of solutions =S=9π+95π+97π=913π
JEE Main 2021 — Mathematics Trigonometry
If n is the number of solutions of the equation 2cosx(4sin(4π+x)sin(4π−x)−1)=1, x∈[0,π] and S is the sum of all these solutions, then the ordered pair (n,S) is :
Held on 1 Sept 2021 · Verified 6 Jul 2026.
(2,98π)
(3,913π)
(2,32π)
(3,35π)
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