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
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)
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If $\cot x=\frac{5}{12}$ for some $x \in\left(\pi, \frac{3 \pi}{2}\right)$, then $\sin 7 x\left(\cos \frac{13 x}{2}+\sin \frac{13 x}{2}\right)+\cos 7 x\left(\cos \frac{13 x}{2}-\sin \frac{13 x}{2}\right)$ is equal to
The value of sin²30° + cos²30° is:
Considering the principal values of inverse trigonometric functions, the value of the expression $\tan \left(2 \sin ^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos ^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :
The number of solutions of 2sin²x + sin²2x = 2 in [0, 2π] is:
The value of $\frac{\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}}$ is equal to
Work through every JEE Main Trigonometry PYQ, year by year.