How do I deal with sines or cosines greater than $1$? When I'm solving trigonometric equations, I occasionally end up with a sine or cosine that's greater than $1$ -- and not on the unit circle. For example, today I had one that was $3 \tan 3x = \sqrt{3}$, which simplified to $\tan 3x = \frac{\sqrt{3}}{3}$, which simplified to $\sin 3x = \sqrt{3}$ and $\cos 3x = 3$. So far as I know, I can't divide this by $3$ (to isolate $x$) until I get angles with a sine of $\sqrt{3}$ and a cosine of $3$, respectively. So, how do I reduce this cosine so that I can find an angle on the unit circle with that cosine? Or, am I doing something very wrong to get this as a cosine in the first place? 
Thanks!
evamvid
 A: As pointed out in the comments, you are doing something like the following:
$$\frac{2}{4} = \frac{1}{2} \qquad \Rightarrow \qquad 2=1\quad \mbox{ and } \quad 4=2.$$
This is very incorrect.
When we are at $\tan 3x = \frac{\sqrt{3}}{3}$, the correct thing to do is find $y$ such that $\tan y = \frac{\sqrt{3}}{3}$ and then let $x=y/3$. On your calculator, you would compute $\arctan(\sqrt{3}/3)$ and set $y=\arctan(\sqrt{3}/3)+\pi n$ for $n=0, \pm 1, \pm 2, \dots$. Remember, there are multiple solutions because $\tan y$ has period $\pi$.
What if you don't want to use a calculator? Note that $\sqrt{3}/3$ is $1/\sqrt{3}$. Now think about a 30-60-90 triangle. Hint: Think about the 30 degree, or $\pi/6$ angle.
A: $$
\tan(3x) = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}
$$
$$
\frac{\sin(3x)}{\cos(3x)} = \frac{1}{\sqrt{3}}
$$
$$
\sqrt{3}\sin(3x) = \cos(3x)
$$
$$
1 = (\sin (3x))^2 + (\cos (3x))^2 = (\sin (3x))^2 + (\sqrt{3}\sin(3x))^2
$$
$$
1 = (\sin (3x))^2 + 3(\sin (3x))^2 = 4(\sin(3x))^2
$$
$$
(\sin(3x))^2 = \frac 1 4
$$
$$
\sin(3x) = \pm\frac 1 2
$$
$$
\cos(3x) = \pm\frac{\sqrt{3}}{2}
$$
(If one of them is "$+$" then so is the other; otherwise they're both "$-$".)
A: Hint: If you are solving $\tan x = a/b$, you have to realize $a/b$ can be written as $(a/c)/(b/c)$ for any nonzero $c$. You have to pick a value of $c$ before you can try to do that. Look at $c= \pm \sqrt{a^2+b^2}$ and see if either of those values gets you further.
