The given equation can be written as:
1−2sin2x+αsinx=2α−7
⇒2sin2x−αsinx+2α−8=0
⇒sinx=2α−4,2(sinx=2)
For at least one solution
−1≤2α−4≤1
⇒α∈[2,6]
Let S be the set of all α∈R such that the equation, cos2x+αsinx=2α−7 has a solution. Then S is equal to:
Held on 12 Apr 2019 · Verified 6 Jul 2026.
[3,7]
[2,6]
[1,4]
R
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.