How to cancel out sine?

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.

To help understand, let's make up a simple problem. (x-1)x = 1/2(x-1). If you divide by (x-1) on both sides to solve, the answer is x = 1/2. But let's multiply it out and solve a different way.

(x-1)x = 1/2(x-1)

x2 - x = 1/2x - 1/2

2(x2 - x) = 2(1/2x - 1/2)

2x2 - 2x = x - 1

2x2 - 3x + 1 = 0

(2x - 1)(x - 1) = 0

(2x - 1) = 0 or (x - 1) = 0

x = 1/2 or x = 1

Plugging those numbers in, we see that both are truly correct answers.

To solve cos x = 1, follow these steps:

Steps 2 and 3 illustrate the different ways that you can write the answers: either as a few within a certain interval, or as all that are possible, with a rule to describe them.

The following example involves a reciprocal function. Your best bet is to begin by using a reciprocal identity and changing the problem.