In the following problem, I first did it using a cancellation of $sin^2\theta$, working shown below, which gave the wrong answer. Having looked at the question again, I saw it could be solved by factoring, working again below.
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.
Correct Solution
$$4\sin\theta = \frac{\sin\theta}{\cos\theta}$$
$$4\sin\theta \cos\theta – \sin\theta = 0$$
$$\sin\theta(4\cos\theta – 1) = 0$$
Rest of working to final answer omitted.
In sum, why is it wrong to cancel as I did first time around – why must these problems be solved by factoring as in the correct solution?
Best Answer
The two issues are:
When cancelling a factor, note that this is only possible when the factor is not zero; but the factor may be zero in the solution to the problem. In this case $\sin\theta = 0$ is a solution. By cancelling you neglected to check that this was a solution.
When squaring, note that this may introduce false solutions, since $a^2=b^2$ whenever $a=b$ or $a=-b$. In this case your answer is that $\cos\theta = \pm\frac 1 4$. But $\cos\theta = {-} \frac 1 4$ is not a true solution.