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Solve sin^4(x)+cos^4(x)=cos4x

User XmlParser
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\sin^4x=(\sin^2x)^2=\left(\frac{1-\cos2x}2\right)^2=\frac{1-2\cos2x+\cos^22x}4

\cos^4x=(\cos^2x)^2=\left(\frac{1+\cos2x}2\right)^2=\frac{1+2\cos2x+\cos^22x}4


\implies \sin^4x+\cos^4x=\frac{2+2\cos^22x}4=\frac{1+\cos^22x}2


\cos^22x=\frac{1+\cos4x}2

\implies\sin^4x+\cos^4x=\frac{1+\frac{1+\cos4x}2}2=\frac{3+\cos4x}4

So the equation reduces to


\frac{3+\cos4x}4=\cos4x

\frac34=\frac34\cos4x

\cos4x=1

We have
\cos\theta=1 whenever
\theta=0+2k\pi=2k\pi, where
k is any integer. This means for this equation we have solutions at


4x=2k\pi\implies x=\frac{k\pi}2
User CocoNess
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