TITLE: How should I go about finding angles at which two trigonometric ratios are equal?
QUESTION [0 upvotes]: Let's say we have an equation, $ \sin x = \cos x $. We know that between $ 0^o \text{and} \ 90^o $, $ \sin $ and $ \cos $ are equal for $ x = 45^o $. This is something we can easily figure because we memorize the trig ratios for angles between 0 and 90. How should I go about finding other angles for which an equality between two ratios hold? 
I found by trial and error that this equation is true for $ x = 225^o $-- but I'd like to if there is a method to do it without trial and error and for all other trigonometric ratios as well (for example, say, $ \tan x = \cot x $).

REPLY [1 votes]: Use that from $$\sin(x)=\cos(x)$$ we get $$\tan(x)=1$$ if $$\cos(x)\neq 0$$

REPLY [1 votes]: Rewrite the equation to have only one trigonometric ratio rather than two
$$\sin x = \cos x \iff 1 = \frac{\sin x}{\cos x} \iff \tan x = 1$$
for $\cos x \neq 0$.
For $n \in \mathbb{Z}$, we have
$$\tan x = 1 \iff x = \begin{cases} \arctan 1 = \frac{\pi}{4}+2\pi n \\ \pi+\arctan 1 = \frac{\pi}{4}+\pi+ 2\pi n \end{cases}$$
which means $x = \frac{\pi}{4}+\pi n$, which is $x = 45°+180°n$ in degrees.
For the other example you gave, we have $\tan x = \cot x$, but $\cot x = \frac{1}{\tan x}$, so rewrite the equation as 
$$\tan x = \frac{1}{\tan x}$$
$$\tan^2 x = 1 \iff \tan x = \pm 1$$
The two conditions give
$$\tan x = 1 \iff x = \frac{\pi}{4}+\pi n$$
$$\tan x = -1 \iff x = \frac{3\pi}{4}+\pi n$$
which can be combined to give $x = \frac{\pi}{4}+\frac{\pi n}{2}$, which is $x = 45°+90°n$ in degrees.
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