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cos-4-x-sin-4-x-cos-2-x-x-




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

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