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Question

Solve 3cos2θ23sinθcosθ3sin2θ=0

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Solution

Given, 3cos2θ3sinθcosθ3sin2θ=03cos2θ3sin2θ=03cos2θsin2θ=0
32cos2θ12sin2θ=0
sin(π32θ)=0
π32θ=nπ2θ=nππ3

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