Solution assignment 08 Trigonometric equations

Solve the following equation for $0\leq{x}\leq{2\pi}$ :

$\sin(\displaystyle\frac{\pi}{2}-x)=\sin(x+\displaystyle\frac{\pi}{3})$

In this case we have:

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

$x=\displaystyle\frac{\pi}{12}$

or:

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

but no value of $x$ can satisfy this equation.
Thus the only solution is:

$x=\displaystyle\frac{\pi}{12}=0.28$