Question Number 86723 by A8;15: last updated on 30/Mar/20
Commented by jagoll last updated on 30/Mar/20
$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{tan}\:\mathrm{x}\:×\:\mathrm{cot}\:\mathrm{x} \\ $$$$\mathrm{cos}\:\mathrm{2x}\:=\:\mathrm{1}\:\Rightarrow\:\begin{cases}{\mathrm{x}\:=\:\mathrm{0}}\\{\mathrm{x}\:=\:\pi}\end{cases} \\ $$