Question Number 154748 by imjagoll last updated on 21/Sep/21
Answered by ARUNG_Brandon_MBU last updated on 21/Sep/21
$$\mathrm{sin}\left(\mathrm{3log}_{\left(\mathrm{2sin}{x}\right)} \sqrt[{\mathrm{3}}]{\pi}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{log}_{\left(\mathrm{2sin}{x}\right)} \pi=\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{log}_{\pi} \left(\mathrm{2sin}{x}\right)=\frac{\mathrm{6}}{\pi}\:\Rightarrow\mathrm{sin}{x}=\frac{\mathrm{1}}{\mathrm{2}}\pi^{\frac{\mathrm{6}}{\pi}} \\ $$$$\Rightarrow{x}=\mathrm{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi^{\frac{\mathrm{6}}{\pi}} \right) \\ $$
Commented by mathdanisur last updated on 21/Sep/21
$$\mathrm{and}\:\:\mathrm{x}=\pi-\mathrm{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\pi^{\frac{\mathrm{6}}{\boldsymbol{\pi}}} \right)+\mathrm{2}\pi{n} \\ $$