How to find all solutions of the equation $\sin x+\cos x=0$ which belong to $(-\pi, \pi)$?

Question

Could you please help me understand and answer this question?

Find all  the  solutions of this equation $$ \sin x+\cos x=0 $$ which belong  to  the interval $(-π; π)$

Progress

Divided by $\cos x$, got $\tan x=-1$.

Answer 1

It's quite straightforward: $$0 = \sin x + \cos x = \frac1{2i}(e^{i x}-e^{-i x})+\frac12(e^{i x}+e^{-i x}),$$ multiplying by $2ie^{i x}$ gives an equivalent form $$0=e^{2i x}-1+ie^{2i x}+i\iff 1-i=e^{2i x}(i+1)\iff e^{2 i x}=\frac{1-i}{i+1}=\frac{-i(i+1)}{i+1}=-i=e^{-i\pi/2},$$ which is true if and only if $2ix = -i\pi/2+2i\pi k\iff x=-\pi/4+\pi k$ for some $k\in\mathbb Z$, so the only solutions inside the interval are $x=-\pi/4$ and $x=3\pi/4$.

Answer 2

Hint: $\sin x + \cos x = 0$ if and only if $\sin x = -\cos x$ if and only if $\tan x = -1$.