Question Number 205220 by hardmath last updated on 13/Mar/24
$$\mathrm{cos}^{\mathrm{4}} \:\mathrm{x}\:−\:\mathrm{sin}^{\mathrm{4}} \:\mathrm{x}\:=\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{x} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:? \\ $$
Answered by Sutrisno last updated on 13/Mar/24
$$\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)\left({cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}\right)={cos}^{\mathrm{2}} {x} \\ $$$${cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}={cos}^{\mathrm{2}} {x} \\ $$$${sinx}=\mathrm{0} \\ $$$${x}={k}.\mathrm{360}° \\ $$$${x}=\mathrm{180}+{k}.\mathrm{360}° \\ $$
Answered by A5T last updated on 13/Mar/24
$${sin}^{\mathrm{2}} {x}=\mathrm{1}−{cos}^{\mathrm{2}} {x}\Rightarrow{sin}^{\mathrm{4}} {x}=\mathrm{1}+{cos}^{\mathrm{4}} {x}−\mathrm{2}{cos}^{\mathrm{2}} {x} \\ $$$$\Rightarrow{cos}^{\mathrm{4}} {x}−{sin}^{\mathrm{4}} {x}=\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}={cos}^{\mathrm{2}} {x}\Rightarrow{cos}^{\mathrm{2}} {x}=\mathrm{1} \\ $$$$\Rightarrow{cos}\left({x}\right)=\underset{−} {+}\mathrm{1}\Rightarrow{x}=\mathrm{360}°{n};\mathrm{180}°+\mathrm{360}{n} \\ $$