Darren Egilsson

My question then, is why is it wrong to cancel the $\sin^2\theta$ term - the algebra seems correct to me?

The Problem

Solve for $\theta$ in the interval $0 \le \theta \le 360$

$$4\sin\theta = \tan\theta$$

My Solution

$$4\sin\theta = \frac{\sin\theta}{\cos\theta}$$ $$4\sin\theta \cos\theta = \sin\theta$$

Squaring, and substituting, using the identity $\cos^2\theta = 1 - \sin^2\theta$

$$16\sin^2\theta(1-\sin^2\theta) = \sin^2\theta$$ $$1-\sin^2\theta = \frac{\sin^2\theta}{16\sin^2\theta}$$ $$1-\sin^2\theta = \frac{1}{16}$$

Rest of working to final answer omitted.

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