Question Number 208686 by hardmath last updated on 21/Jun/24
$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{2x}\:=\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2x}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:? \\ $$
Answered by MM42 last updated on 21/Jun/24
$${cos}^{\mathrm{2}} \mathrm{2}{x}−{sin}^{\mathrm{2}} \mathrm{2}{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${cos}\mathrm{4}{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow\mathrm{4}{x}=\mathrm{2}{k}\pi\pm\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow{x}=\frac{{k}\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{24}}\:\:\checkmark \\ $$$$ \\ $$
Commented by hardmath last updated on 28/Jun/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$