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?

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    $\begingroup$ When squaring be aware that $a^2=b^2$ whenever $a=b$ or $a={-b}$ $\endgroup$ Sep 1, 2015 at 4:28
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    $\begingroup$ When canceling, note that the term canceled can be zero. $\endgroup$ Sep 1, 2015 at 4:31
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    $\begingroup$ Specifically, the decision to divide through as you did in order to "cancel" the $ \ \sin^2 \ \theta \ $ factor is based on the assumption that it is not zero. It is "safe" to do this in an equation where the factor that is being divided is known (say, from conditions in the problem) not to be zero. But if that factor could be zero, then you are choosing to ignore it by doing the division. (You would then need to consider separately the possibility of what happens if that factor is equal to zero.) To avoid this complication, it is preferred to use the method in the other solution. $\endgroup$ Sep 1, 2015 at 4:36
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    $\begingroup$ The equation $4x=x^2$ has two solutions, while (after cancelling $x$) the equation $4=x$ has only one $\endgroup$
    – Henry
    Sep 1, 2015 at 10:07
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    $\begingroup$ Also note that you could use your method to solve the problem, but only if you also take into account the spurious solutions that your method may generate as well as the valid solutions your method may leave behind. In most cases, that ends up being more tedious work that also can lead to more mistakes. So your method isn't 100% wrong, it's just not the best method in this case. $\endgroup$ Sep 1, 2015 at 12:11

5 Answers 5


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.

  • $\begingroup$ yes but WHY IS Cosx = -1/4 not a true solution? the cosine functon can take any value between +1 and -1 $\endgroup$
    – user273942
    Sep 25, 2015 at 6:13
  • $\begingroup$ @NeilMcBain Because it does not satisfy the initial equation. $4\sin(\arccos (-\tfrac 14)) \neq \tan(\arccos(-\tfrac 14))$ $\endgroup$ Sep 25, 2015 at 8:41

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.


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 $\theta \in \mathbb{R}$


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 $

Case 2: $sin \space θ \neq 0$

$4 \space cos \space θ = 1 $
$ \displaystyle cos \space θ = \frac{1}{4}$


If you like this, then always remember to break into cases.


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:

$cos(\theta) = \frac{1}{4}$


$acos(\frac{1}{4}) = \theta$

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    $\begingroup$ ... aaaand you get one of the two errors that OP had. - For exactly the same reason. $\endgroup$
    – Taemyr
    Sep 1, 2015 at 13:40
  • $\begingroup$ @Taemyr, oops. I guess I should have checked the possibility that a equaling zero solves the equation before cancelling it. $\endgroup$ Sep 1, 2015 at 14:59

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