Question Number 18723 by mondodotto@gmail.com last updated on 28/Jul/17
Answered by Tinkutara last updated on 29/Jul/17
$$\mathrm{2sin2}\theta\:+\:\mathrm{15cos2}\theta\:=\:\mathrm{10} \\ $$$$\frac{\mathrm{2sin2}\theta}{\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{15cos2}\theta}{\sqrt{\mathrm{229}}}\:=\:\frac{\mathrm{10}}{\sqrt{\mathrm{229}}} \\ $$$$\mathrm{cos}\left(\mathrm{2}\theta\:−\:\alpha\right)\:=\:\frac{\mathrm{10}}{\sqrt{\mathrm{229}}}\:,\:\mathrm{where}\:\alpha\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\sqrt{\mathrm{229}}}\right) \\ $$$$\theta\:=\:{n}\pi\:\pm\:\frac{\mathrm{5}}{\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\sqrt{\mathrm{229}}}\right) \\ $$