JEE Main 2022Trigonometric EquationsTrigonometric EquationsHardQuestionThe sum of all values of θ∈[0,2π]\theta \in[0,2 \pi]θ∈[0,2π] satisfying 2sin2θ=cos2θ2 \sin ^2 \theta=\cos 2 \theta2sin2θ=cos2θ and 2cos2θ=3sinθ2 \cos ^2 \theta=3 \sin \theta2cos2θ=3sinθ isOptionsAπ\piπB5π6\frac{5 \pi}{6}65πCπ2\frac{\pi}{2}2πD4π4 \pi4πCheck AnswerHide SolutionSolution2sin2θ=cos2θ2sin2θ=1−2sin2θ4sin2θ=1sin2θ=14sinθ=±122cos2θ=3sinθ2−2sin2θ+3sinθ−2=0(2sinθ−1)(2sinθ−2)=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}2sin2θ=cos2θ2sin2θ=1−2sin2θ4sin2θ=1sin2θ=41sinθ=±212cos2θ=3sinθ2−2sin2θ+3sinθ−2=0(2sinθ−1)(2sinθ−2)=0sinθ=21 so common equation which satisfy both equations is sinθ=12\sin \theta=\frac{1}{2}sinθ=21 θ=π6,5π6(θ∈[0,2π])\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])θ=6π,65π(θ∈[0,2π]) Sum =π\text { Sum }=\pi Sum =π