Solution assignment 04 Trigonometric equations

Solve the following equation:

$\sin(x+\displaystyle\frac{\pi}{4})=\sin(\displaystyle\frac{\pi}{3})$

The first three solutions are (using the fact that the $\sin$ -function has a period $2\pi$ ):

$x+\displaystyle\frac{\pi}{4}=\displaystyle\frac{\pi}{3}+2k\pi, k=0, \pm1, \pm2, ...$

$x=\displaystyle\frac{1}{12}\pi+2k\pi, k=0, \pm1, \pm2, ...$

The second series solutions are:

$x+\displaystyle\frac{\pi}{4}=\pi-\displaystyle\frac{\pi}{3}+2k\pi, k=0, \pm1, \pm2, ...$

$x=\displaystyle\frac{5}{12}\pi+2k\pi, k=0, \pm1, \pm2, ...$