<s>[INST] Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation 
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side 
Thus, I had 
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
I got  $$s + 13 = 4 + 4\sqrt{7-s}+  7 – s$$
I then combined like terms 
$$2s + 2= 4 \sqrt{7-s}$$
I’m stuck at this point. Does anyone have an idea how to solve this equation?
 [/INST] Clearly $7\ge s\ge-13\iff7-(-3)\ge s+3\ge-13-(-3)$
WLOG $s+3=10\cos 2y$ where $0\le2y\le\pi$
$\implies\sqrt{s+13}=\sqrt{10(1+\cos2y)}=2\sqrt5\cos y,\sqrt{7-s}=2\sqrt5\sin y$
$\implies2\sqrt5\cos y-2\sqrt5\sin y=2$
Squaring we get $$20(1-\sin2y)=4\iff\sin2y=\dfrac45\implies\cos2y=\pm\sqrt{1-\sin^22y}=\cdots$$
Check which values of $s=10\cos2y-3$ satisfies the given equation.
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