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Question Number 205220 by hardmath last updated on 13/Mar/24

$${\cos }^4\mathrm{x}-{\sin }^4\mathrm{x}={\cos }^2\mathrm{x}$$$$\Rightarrow \mathrm{x}=?$$

Answered by Sutrisno last updated on 13/Mar/24

$$({cos}^{2}x+{sin}^{2}x)({cos}^{2}x-{sin}^{2}x)={cos}^{2}x$$$${cos}^{2}x-{sin}^{2}x={cos}^{2}x$$$$sinx=0$$$$x=k.360°$$$$x=180+k.360°$$

Answered by A5T last updated on 13/Mar/24

$${sin}^{2}x=1-{cos}^{2}x\Rightarrow {sin}^{4}x=1+{cos}^{4}x-2{cos}^{2}x$$$$\Rightarrow {cos}^{4}x-{sin}^{4}x=2{cos}^{2}x-1={cos}^{2}x\Rightarrow {cos}^{2}x=1$$$$\Rightarrow cos\left(x\right)=\underset{-}{+}1\Rightarrow x=360°n;180°+360n$$

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