Solve equation $ \cos x+\sin x=0$

Note that $$\cos x + \sin x = 0 \iff \cos x = -\sin x$$

Now, $\cos x$ cannot equal zero, since if it did, $\sin x = -1$ or $\sin x = 1$, in which case the given equation isn't satisfied.

So we can divide by $\cos x$ to get $$1 = \dfrac{-\sin x}{\cos x} = -\tan x \iff \tan x = -1$$

Solving for $x$ gives us the values $x = \dfrac {3\pi}4 + k\pi$, where $k$ is any integer.

Another way to solve this is to write $\cos x = (e^{ix}+e^{-ix})/2$ and $\sin x = (e^{ix}-e^{-ix})/2i$. The equation simplifies in a couple of easy steps to $e^{2ix}= e^{-\pi i/2}$. This is equivalent to $2x= -\pi /2 + 2\pi n$, so $x= -\pi /4 + \pi n$ for integral $n$.

Following where you got stuck and squaring both sides and you obtain

$$\sin^2 x+\cos^2x+2\sin x \cos x=0 \Rightarrow 1+\sin2x=0$$

Using $\sin 2x = 2\sin x \cos x$.

This means that

$$\sin 2x = -1$$ and hence $$2x = \frac {3\pi}{2}+2k\pi \Rightarrow x=\frac {3\pi}{4}+k\pi$$