Skip to main content
Back to Trigonometric Equations
JEE Main 2022
Trigonometric Equations
Trigonometric Equations
Hard

Question

The sum of all values of θ[0,2π]\theta \in[0,2 \pi] satisfying 2sin2θ=cos2θ2 \sin ^2 \theta=\cos 2 \theta and 2cos2θ=3sinθ2 \cos ^2 \theta=3 \sin \theta is

Options

Solution

2sin2θ=cos2θ2sin2θ=12sin2θ4sin2θ=1sin2θ=14sinθ=±122cos2θ=3sinθ22sin2θ+3sinθ2=0(2sinθ1)(2sinθ2)=0sinθ=12\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} so common equation which satisfy both equations is sinθ=12\sin \theta=\frac{1}{2} θ=π6,5π6(θ[0,2π])\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])  Sum =π\text { Sum }=\pi

Practice More Trigonometric Equations Questions

View All Questions