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

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

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

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

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

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