Question Number 164511 by HongKing last updated on 18/Jan/22
$$\mathrm{sin}\centerdot\left(\pi\:\mathrm{sin}\:\mathrm{x}\right)\:-\:\mathrm{cos}\centerdot\left(\pi\:\mathrm{sin}\:\mathrm{x}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$
Answered by mindispower last updated on 18/Jan/22
$${sin}\left({a}\right)−{cos}\left({a}\right)=\sqrt{\mathrm{2}}{sin}\left({a}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Leftrightarrow\:{withea}=\pi{sin}\left({x}\right) \\ $$$${sin}\left({a}−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Leftrightarrow{a}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi,\frac{\mathrm{3}\pi}{\mathrm{4}}+\mathrm{2}{k}\pi \\ $$$${sin}\left({x}\right)=\mathrm{1}+\mathrm{2}{k}\in\left\{\mathrm{1},−\mathrm{1}\right\} \\ $$$${x}=\frac{\pi}{\mathrm{2}}+{k}\pi \\ $$$$\pi{sin}\left({x}\right)=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi \\ $$$${sin}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}{k},\:\:\Rightarrow{k}=\mathrm{0} \\ $$$${sin}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}},{x}\in\left\{\frac{\pi}{\mathrm{6}}+\mathrm{2}{k}\pi,\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2}{k}\pi\right\} \\ $$$$ \\ $$
Commented by HongKing last updated on 20/Jan/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$