How many solutions does a trigonometric function have $0\le x \le 2\pi$? 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  $
b)$(\tan(2x))^2=1$
Thanks in advance.
 A: 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:
$$\sin3x=-\frac14\iff3x=\arcsin\left(-\frac14\right)+2k\pi\ldots\ldots\text{etc.}$$
A: 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.
