Solve the equation $\displaystyle 2sin(x-45^0) = 1 $ for $\displaystyle 0^0<x<360^0$
Solve it the way you normally would
$\displaystyle 2\sin{(x - 45^{\circ})} = 1$
$\displaystyle \sin{(x - 45^{\circ})} = \frac{1}{2}$
$\displaystyle x - 45^{\circ} = \left\{30^{\circ}, 180^{\circ} - 30^{\circ}\right\}$
$\displaystyle x - 45^{\circ} = \left\{30^{\circ}, 150^{\circ}\right\}$
$\displaystyle x = \left\{30^{\circ} + 45^{\circ}, 150^{\circ} + 45^{\circ}\right\}$
$\displaystyle x = \left\{75^{\circ}, 195^{\circ}\right\}$.