$\begin{aligned}
& 2 \sin ^2 \theta=\cos 2 \theta \
& 2 \sin ^2 \theta=1-2 \sin ^2 \theta \
& 4 \sin ^2 \theta=1 \
& \sin ^2 \theta=\frac{1}{4} \
& \sin \theta= \pm \frac{1}{2} \
& 2 \cos ^2 \theta=3 \sin \theta \
& 2-2 \sin ^2 \theta+3 \sin \theta-2=0 \
& (2 \sin \theta-1)(2 \sin \theta-2)=0 \
& \sin \theta=\frac{1}{2}
\end{aligned}socommonequationwhichsatisfybothequationsis\sin \theta=\frac{1}{2}\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])Sum=\pi$
The sum of all values of θ ∈[0,2 π] satisfying 2 sin ^2 θ=cos 2 θ and 2 cos ^2 θ=3 sin θ is
JEE Main 2025 — Mathematics Trigonometry
The sum of all values of θ∈[0,2π] satisfying 2sin2θ=cos2θ and 2cos2θ=3sinθ is
Held on 22 Jan 2025 · Verified 6 Jul 2026.
Options
- A
4π
- B
65π
- C
π
- D
2π
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