Solve: sin(x+11pi/6)-sin(x-11pi/6)=1

$\sin(x +\frac{11\pi}{6})\,-\,\sin(x -\frac{11\pi}{6})\,=\,1$

We can use the sum of angles formula for sine:

$\sin(\alpha \pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\~\\ \ \\ {\color{BlueViolet} \sin(x +\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}+\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\\~\\ {\color{Fuchsia} \sin(x -\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}-\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}$

So if...

${\color{BlueViolet}\sin(x +\frac{11\pi}{6})}\,-\,{\color{Fuchsia}\sin(x -\frac{11\pi}{6})}\,=\,1\\~\\ [\,{\color{BlueViolet}(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\,]\,-\,[\,{\color{Fuchsia}(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}\,]\,=\,1\\~\\ (\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)-(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)\,=\,1\\~\\ 2(\cos x)(-\frac12)\,=\,1\\~\\ - \cos x\,=\,1\\~\\ \cos x\,=\,-1\\~\\ x\,=\,\pi+2\pi n\quad \text{, where }\ n\ \text{ is an integer.}$      then...