If $\sin \alpha+\cos \alpha=-\frac{\sqrt 7}{2}$, then $\alpha$ is the angle of which quadrant? Question:

If $$\sin \alpha+\cos \alpha=-\frac{\sqrt 7}{2}$$ then $\alpha$ is the angle of which quadrant?

My attempts:
$$\sqrt2\sin\left(\frac \pi 4+\alpha\right)=-\frac{\sqrt {7}}{2}$$
$$\sin\left(\frac \pi 4+\alpha\right)=-\frac{\sqrt {14}}{4}$$
Here, I am stuck. I can not continue.
 A: $$\sin \left(\frac \pi 4+\alpha\right)=-\frac{\sqrt{14}}{4}$$
$$\frac \pi 4+\alpha=2k\pi-\sin^{-1}\left(\frac{\sqrt{14}}{4}\right)\ \ \text{OR}\ \  \ \  \frac \pi 4+\alpha=2k\pi-\pi+\sin^{-1}\left(\frac{\sqrt{14}}{4}\right)$$
$$\alpha=\frac{(8k-1)\pi}{4}-\sin^{-1}\left(\frac{\sqrt{14}}{4}\right)\ \ \text{OR}\ \  \ \  \alpha=\frac{(8k-5)\pi}{4}+\sin^{-1}\left(\frac{\sqrt{14}}{4}\right)$$
$$\alpha\in\{\frac{(8k-1)\pi}{4}-\sin^{-1}\left(\frac{\sqrt{14}}{4}\right)\} \cup\{ \frac{(8k-5)\pi}{4}+\sin^{-1}\left(\frac{\sqrt{14}}{4}\right)\}$$
Where, $k$ is any integer i.e. $k=0, \pm1, \pm2, \pm3, \ldots$
A: The minimum value of both $\sin\alpha$ and $\cos\alpha$ is $-1$.
Hence, if either $\sin\alpha$ or $\cos\alpha$ was positive, they could not sum to a value less than $-1$.
But they sum to $-\frac{\sqrt7}{2}$, which is less than $-1$. 
Hence, both $\sin\alpha$ and $\cos\alpha$ are negative, which means that $\alpha$ is in the $3$rd quadrant.
A: Generally $\sin$ is negative from $(2n-1)π$ to $2nπ$.
So we have:
$$(2n-1)π<\alpha+\fracπ4<2nπ$$
$$\implies(2n-1)\pi-\fracπ4<\alpha<2nπ-\fracπ4$$
Now you can put different integer values of $n$ to get different ranges for $\alpha$.
A: The other answers and comments give you an effective way of solving the problem, which I advise you use instead since it is simpler and, more importantly, faster. But for completeness, it is also possible to continue from where you left off with:
$$\begin{align}
\sin\left(\frac{\pi}{4}+\alpha\right)&=\frac{\sqrt{14}}{4}\\
\alpha &= \sin^{-1}\frac{\sqrt{14}}{4}-\frac\pi4 +2\pi n
\\\text{or}\quad \alpha &= \pi - \sin^{-1}\frac{\sqrt{14}}{4} -\frac\pi4+2\pi n
\end{align}$$
You can solve your equation for $\alpha$ by taking the arcsine of both sides. But recall that inverting sine gives us two principal solutions: both $\sin^{-1}x$ and $\pi-\sin^{-1}x$. Also, any solution plus $2\pi n$, where $n$ is an integer, is a solution too. Taking this into account, you now have all solutions.
Pick your favourite integer for $n$ and compute the value of $\alpha$. You should find it is invariably in the third quadrant.
A: Substitute $\sin a =\frac{2\tan\frac a2}{1+\tan^2\frac a2}$ and $\cos a =\frac{1-\tan^2\frac a2}{1+\tan^2\frac a2}$ in $\sin \alpha+\cos \alpha=-\frac{\sqrt 7}{2}$,
$$(\sqrt7-2)\tan^2\frac a2 +4\tan \frac a2 +(2+\sqrt 7)=0$$
Solve to get $\tan \frac a2= -(2+\sqrt7), -\frac{2+\sqrt7}3<-1$, i.e. $-\frac\pi2<\frac a2 +\pi n< -\frac\pi4$, or
$$-\pi<a+2\pi n<  -\frac\pi2$$
Thus, $a$ is in the 3rd quadrant.
A: Just draw a picture.  Quadrant I is a "positive" quadrant.  $\sin x, \cos x \ge 0$ (but not both at once) so $\sin x+ \cos x > 0$.  (A little unneeded analysis can see that $\sin' x = \cos x$ and $\cos'x =-\sin x$ so the sum will increase from $1$ at $x=0$ to a maximum of $\sqrt 2$ and $x = \frac \pi 4$ and the decrease back to $1$ and $\frac \pi 2$... but we don't need to do that... it's enough to note it the sum is positive.)
Quadrants II and IV are "mixed" quadrants.  One of the $\sin x, \cos x$ will be positive (or zero) and the other will be negative (or $0$) and the sum will be the difference between numbers between $0$ and $1$ so the result will be between $1$ and $-1$.  As $-\frac {\sqrt 7}2<- 1$ these quadrants won't work
That leaves Quadrant III-- a "negative" quadrant.  It is exactly like Quadrant I but negative.  This is the only quadrant left.  As $|\sin x|, |\cos x| \le 1$ and $\sin \alpha + \cos \alpha < -1$ we must have both $\sin \alpha$ and $\cos \alpha$ negative and Quadrant III is the only quadrant that is possible.
Of course that doesn't mean that there IS a solution possible.  For example $\sin x + \cos x = -2$ has no solution in any quadrant.  So to show there is a solution, I guess we do need to do the "unneeded" analysis w did in Quadrant I.  Note that $-1 >-\frac{7}2 > -\sqrt 2$ so  that indeed there will be two $\alpha$s solutions in this quadrant.
.......
But let's be differen for the sake of being different.
If $\sin \alpha + \cos \alpha = -\frac {\sqrt 7}2$ then
$(\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha +2\cos a\sin \alpha + \cos^2 \alpha =1+2\cos\alpha\sin \alpha = \frac 74$
Is $\cos\alpha\sin\alpha = \frac 38$.
And so $\cos \alpha$ and $\sin \alpha$ must both be the same sign.  And they can't both be positive as the sum is negative.
So quadrant III.
(Note the least non-negative value $\cos\alpha\sin \alpha$ can be is $0$ and the maximum in can be is $\frac {\sqrt{2}}2\cdot \frac {\sqrt{2}}2 = \frac 14$ so this all fits.
