So I have the following equation:
$$ \frac{\cos(x)}{\sin(x)}=\frac{3}{\sqrt3}$$
and when solving it for $x$ (in the range of $[0 - \pi]$ ), I know the solution is $x=\pi/6$.
However I don't understand how this is done. What steps do I need to take to find $x$ for this and similar equation?
HINT:
$$\tan(x) = \frac{\sin x}{\cos x} \iff \frac{1}{\tan x} = \frac{\cos x}{\sin x}$$
This should be more straightforward to solve
Hint: you can also write $$\tan(x)=\frac{\sqrt{3}}{3}$$
Equivalently $$ \tan(x)=\frac{\sin(x)}{\cos(x)}=\frac{\sqrt{3}}{3}. $$ The tangent function is strictly increasing in $(-\pi,\pi)$ with image from $-\infty$ to $+\infty$, hence the solution exists and is unique. $x=\pi/6$ is the solution.
$\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt3}{3}$
$\tan x = \frac1{\sqrt3}$
$\tan x = \tan \frac{π}{6}$
$x = \frac{π}{6}$
Or
$ \frac{\cos x}{\sin x} = \frac3{\sqrt3}$
$\cot x = \sqrt3$
$\cot x = \cot \frac{π}{6}$
$x = \frac{π}{6}$