Question

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

Answer 1

`\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`