Simple Trig Equations - Why is it Wrong to Cancel Trig Terms? 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?
 A: The problem wants you to find all values of $\theta$ that satisfy the equation. Cancellation makes it easier to find some of the values, but by cancelling you also lose solutions. Factoring ensures that you retain the information needed to find all the solutions.
A: Notice that if $\theta = \pi n$, $\{n \in \mathbb{Z}\}$ then you will have $1 = 0$.
Remember that when we "cancel" what we're really doing is saying this term over this term = 1.  $\frac{\sin\theta}{\sin\theta} \neq 1 $ for $\theta \in \mathbb{R}$
A: 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.
A: I differ here slightly from most of the answers. I personally find that if you are careful dividing through can be a shorter strategy than factoring. Start from the expression below, and divide by $sin \space θ$:  
$4 \space sin \space θ \space cos \space θ = sin \space θ$  
Case 1: $sin \space θ = 0  $
Solve.
Case 2: $sin \space θ \neq 0$  
$4 \space cos \space θ = 1  $
$ \displaystyle cos \space θ = \frac{1}{4}$
Solve.
If you like this, then always remember to break into cases.
A: Here's a simpler way to get your answer.  Remember what $sin(\theta)$ and $tan(\theta)$ actually represent on a right triangle.  For a given right triangle, let $a$ be the length of the adjacent side, $b$ be the length of the opposite side, and $c$ be the length of the hypotenuse.  This means:
$sin(\theta) = \frac{a}{c}$ and $tan(\theta) = \frac{a}{b}$
Substituting, your equation becomes:
$4\frac{a}{c} = \frac{a}{b}$
Cancelling the $a$ and rearranging gives:
$\frac{b}{c} = \frac{1}{4}$
Using the definition of cosine gives:
$cos(\theta) = \frac{1}{4}$
And
$acos(\frac{1}{4}) = \theta$
