(4−3)sinx−23cos2x=1+3−4,x∈[−2π,25π]⇒(4−3)sinx−23(1−sin2x)=2(1−3)⇒23sin2x+4sinx−3sinx−2=0⇒(2sinx−1)(3sinx+2)=0⇒sinx=21∴ Number of solution =5
The number of solutions of equation (4-√3) sin x -2 √3 cos ^2 x=- 4 1+√3, x ∈ [-2 π, 5 π 2 ] is
JEE Main 2025 — Mathematics Trigonometry
The number of solutions of equation (4−3)sinx −23cos2x=−1+34,x∈[−2π,25π] is
Held on 3 Apr 2025 · Verified 6 Jul 2026.
Options
- A
4
- B
3
- C
6
- D
5
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