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

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

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

• Also in general, $x$ could be $-1$ and $y$ an even integer. – Minus One-Twelfth Apr 6 at 12:30
• I understand that y can be 0 with the logaritms law...but how do we solve for x when x=1? – manusakthi natarajan Apr 6 at 12:35
• @manusakthinatarajan $x^y=1\implies|x|^y=1$, now take the logarithm. $y\ln|x|=0\implies y=0\vee|x|=1$ – Shubham Johri Apr 6 at 12:40

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 $$\log 1 = 0$$.

• Thank you. It answers my question. – manusakthi natarajan Apr 6 at 12:45