Question Number 164434 by mathlove last updated on 17/Jan/22
Answered by TheSupreme last updated on 17/Jan/22
$$\mathrm{2}{f}\left({sin}\left({x}\right)\right)−{f}\left({cos}\left({x}\right)\right)=\mathrm{3}{x} \\ $$$$\mathrm{2}{f}\left({cos}\left({x}\right)\right)−{f}\left({sin}\left({x}\right)\right)=\mathrm{3}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\mathrm{3}{f}\left({sin}\left({x}\right)\right)=\mathrm{6}{x}−\mathrm{3}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$${f}\left({sin}\left({x}\right)\right)=\mathrm{9}{x}−\frac{\mathrm{3}}{\mathrm{2}}\pi \\ $$$${f}\left({t}\right)=\mathrm{9}{arcsin}\left({x}\right)−\frac{\mathrm{3}}{\mathrm{2}}\pi+\mathrm{2}{k}\pi \\ $$$${f}\left({tan}\left({x}\right)\right)=\mathrm{9}{arcsin}\left({tan}\left({x}\right)\right)−\frac{\mathrm{3}}{\mathrm{2}}\pi+\mathrm{2}{k}\pi \\ $$$${f}\left({sin}\left(\mathrm{2}{x}\right)\right)=\mathrm{8}{x}−\frac{\mathrm{3}}{\mathrm{2}}\pi+\mathrm{2}{k}\pi \\ $$$$ \\ $$