$\alpha$ can be the angle of which quadrants? Question:

If $$\sin \alpha+\cos \alpha<0$$ then $\alpha$ can be the angle of which quadrants?

My attempts:
$$\sin \alpha+\cos \alpha<0 \Longrightarrow \sin
\left(\frac{\pi}{4}+\alpha\right)<0$$
So I can choose $\left(\frac{ \pi}{4}+\alpha\right) \in \mathrm {III} \thinspace \text{quadrant}$ but $\alpha \in \mathrm {II} \thinspace \text{quadrant}$
Then I can choose $\left(\frac{ \pi}{4}+\alpha\right) \in \mathrm {IIII} \thinspace \text{quadrant}$ but $\alpha \in \mathrm {III} \thinspace \text{quadrant}$
So,  $x\in \mathrm {II, III} \thinspace \text{quadrants}$
Is my solution correct?
 A: A generalized solution for $\alpha$ can be obtained as follows
$$\sin\alpha+\cos\alpha<0$$
$$\sin\left(\alpha+\frac{\pi}{4}\right)<0\implies (2k-1)\pi<\left(\alpha+\frac{\pi}{4}\right)< 2k\pi$$
$$\frac{(8k-5)\pi}{4}<\alpha<\frac{(8k-1)\pi}{4}$$
$$\color{blue}{\alpha\in\left(\frac{(8k-5)\pi}{4}, \frac{(8k-1)\pi}{4}\right)}$$
Where, $k$ is any integer i.e. $k=0, \pm1, \pm2, \pm3,\ldots$
A: Yeah your solution was going to be absolutely correct but u missed it in the last couple of lines.
I think the following solution would be more easy to understand:
sin a + cos a < 0
=> sin a < -cos a
=>tan a < -1
This means that tan a will be always negative.
And this happens in the II and IV quadrant.
Hence, the answer is II or IV quadrant.
PS-I ignored the fact that tan a should be also less than -1 because that would be true for a part of the II and IV quadrant (more precisely, if a diagonal is drawn throught the origin to divide those quadrants equally then a should be in either upper half of the second quadrant or lower half of the IV quadrant)
A: Consider $f(\alpha)=\sin(\alpha)+\cos(\alpha)$.  Its zeroes between $0$ and $2\pi$ are easily calulated to $\alpha=3\pi/4$ and $\alpha=7\pi/4$.  As $f$ is continuous it can't change sign on any of the intervals $]0,3\pi/4[$, $]3\pi/4,7\pi/4[$ and $]7\pi/4,2\pi[$. Thus you'll find out that $f$ is negative only on $]3\pi/4,7\pi/4[$.
