How do I algebraically determine how many solutions a trigonometric solution have $0\leq x\leq 2\pi$? So far I have been drawing graphs for each question and counted the solutions but I want a way to do this algebraically. Many people tell me to use the double angle identities but I haven't learned it yet. I know that the period of $\sin$ and $\cos$ are $2\pi$ and $\tan$ is $\pi$. One of the questions on my test was like..

Determine how many solutions do the following equations have for $0\leq x\leq 2\pi$.

a) $\sin(3x)=-1/4 $


Thanks in advance.

  • $\begingroup$ For the $\sin$ you can write $0\leq 3x\leq 6\pi$ and since $2\pi$ is a period the equation has the same number of solutions from $0$ to $2\pi$ as it has from $2\pi$ to $4\pi$ as it has from $4\pi$ to $6\pi$. Also is it $\tan((2x)^2)$ or $(\tan(2x))^2$,I assume the former but I'm not sure. $\endgroup$
    – kingW3
    Commented Jan 28, 2017 at 22:19
  • $\begingroup$ $0 \le x \le 2\pi \iff 0 \le 3x \le 6\pi$. So $\sin (3x) = -1/4$ wiill have 3 times as many solutions for $0 \le x \le 2\pi$ as $\sin (y) = -1/4$ will have for $0 \le y \le 2\pi$. $\endgroup$
    – fleablood
    Commented Jan 29, 2017 at 0:08

2 Answers 2


I do one, you do the other:

$$\tan^22x=1\iff \tan 2x=\pm1\iff 2x=\pm\frac\pi4+k\pi\;,\;\;k\in\Bbb Z\iff$$

$$\iff x=\pm\frac\pi8+k\frac\pi2\;,\;\;k\in\Bbb Z$$

Hint for the other:



If $0\le x \le 2\pi$ then $0 \le 3x \le 6 \pi$

$\sin (\theta) = -1/4$ will have two solutions in $0 \le \theta \le 2\pi$[$*$] so it will have two solutions in $2\pi \le \theta \le 4\pi$ and two solution is $4\pi \le \theta \le 6\pi$. So there are six solutions for $0 \le \theta = 3x \le 6\pi$ or in other words $0 \le x \le 2\pi$.

$\tan (\theta)^2 = 1 \implies \tan (\theta) = \pm 1$.

$\tan(\theta) = 1 \implies \sin \theta = \cos \theta$ has 2 solution in $0\le \theta \le 2\pi$. (At $\theta = \pi/4$ and $\theta = \pi + \pi/4$).

$\tan(\theta) = -1 \implies \sin \theta = -\cos \theta$ has 2 solutions as well. (At $\pi - \pi/4$ and $2\pi - \pi/4$).

So $\tan (\theta)^2 = 1$ has 4 solutions in $0 \le \theta \le 2\pi$.

If $\theta = 2x$ and $0 \le x \le 2\pi$ then $0 \le 2x = \theta \le 4\pi$ and so there are $8$ solutions for $0 \le 2x = \theta \le 4\pi$.

[$*$] Note: we don't have to actually solve the equation. We know there will be one solution in each the third and fourth quadrant.


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