Solve a given trigonometric equation Solve the following equation:
$$\sin x + \sin \left( x + \frac{7\pi}{24} \right) = \frac{\sqrt{2 - \sqrt 2}}{2} + \frac{\sqrt 6 + \sqrt 2}{4}$$
So far, I found out that $\frac{\sqrt{2 - \sqrt 2}}{2} = \sin \frac{\pi}{8}$ and $\frac{\sqrt 6 + \sqrt 2}{4} = \sin \frac{7\pi}{12}$.
Thank you!
 A: Hint
Expand the second $\sin$, to write your 
equation as
$$A\cos(x)+B\sin(x)=1$$
or
$$\cos(x+\alpha)=\frac{1}{\sqrt{A^2+B^2}}$$

second approach

the equation is
$$
\sin(x)+\sin(x+\frac{7\pi}{24})=$$  
$$\sin( \frac{\pi}{8})+\sin(\frac{7\pi}{12}  )=$$
$$\sin(\frac{\pi}{8})+\sin\left(\pi-(\frac{\pi}{8}+\frac{7\pi}{24})\right)=$$
$$\sin(\frac{\pi}{8})+\sin(\frac{\pi}{8}+\frac{7\pi}{24})$$
A: Let $f(x)=\sin x +\sin (x+7\pi /24).$
We have $$f(\pi /8)=\sin (\pi /8)+\sin (\pi /8+7\pi /24)=\sin (\pi /8)+\sin (10\pi /24)=$$ $$=\sin (\pi /8)+\sin (14\pi /24)=\sin (\pi /8)+\sin (7\pi /12)$$ which is the RHS of your equation.
Now $f'(x)=\cos x +\cos (x+7\pi /24).$ 
So $f(x)$ is strictly increasing for $-\pi /2-7\pi /48\leq x\leq \pi /2-7\pi /48.$
And $f(x)$ is strictly decreasing for $\pi/2 -7\pi /48\leq x< 3\pi/2-7\pi /48.$
So there are at most 2 values of $x$ in $[-\pi/2-7\pi /48,\;\; 3\pi /2-7\pi /48)$ for which $f(x)=f(\pi /8).$ And there do exist 2 values: $x=\pi /8$ and $x=\pi-\pi /8=7\pi /8.$
So $x=\pi /8 +2\pi n$ or $x=7\pi /8+2\pi n$ for some (any) $n\in \mathbb Z.$
