Solving $\sin (x )^{\sin(2x) - \cos(2x)} = 1$ where $0^{\circ}<x<360^{\circ}$.

Question

$$\sin (x )^{\sin(2x) - \cos(2x)} = 1,\quad 0^{\circ}<x<360^{\circ}.$$

Answers are given but working is not found - please help me with this question. Thank you

Answer 1

Hint: If $x^y=1$ then $x=1$ or $y=0$ AND $x\ne0$. This follows from taking logs both sides: $$y\ln{(x)}=0$$

Answer 2

Hint: After taking logarithms, note that

$$(\sin 2x - \cos 2x) \cdot \log (\sin x) = \log 1$$

Thus $$(\sin 2x - \cos 2x) = 0$$ or $$\log (\sin x) = 0$$

as $\log 1 = 0$.