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Exercise 2.2 · Q27

Q.If 2sin⁡2θ+3sin⁡θ=02\sin^2\theta + 3\sin\theta = 0, find the permissible values of cos⁡θ\cos\theta.

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Step 1. 2sin⁡2θ+3sin⁡θ=0⇒sin⁡θ(2sin⁡θ+3)=02\sin^2\theta+3\sin\theta=0 \Rightarrow \sin\theta(2\sin\theta+3)=0.

Step 2. Either sin⁡θ=0\sin\theta=0, or sin⁡θ=−32\sin\theta=-\tfrac32 — rejected, since sinθ can never be less than −1. …

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