Respuesta :

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[tex]\sin^4x=(\sin^2x)^2=\left(\dfrac{1-\cos2x}2\right)^2=\dfrac{1-2\cos2x+\cos^22x}4[/tex]
[tex]\cos^4x=(\cos^2x)^2=\left(\dfrac{1+\cos2x}2\right)^2=\dfrac{1+2\cos2x+\cos^22x}4[/tex]

[tex]\implies \sin^4x+\cos^4x=\dfrac{2+2\cos^22x}4=\dfrac{1+\cos^22x}2[/tex]

[tex]\cos^22x=\dfrac{1+\cos4x}2[/tex]
[tex]\implies\sin^4x+\cos^4x=\dfrac{1+\frac{1+\cos4x}2}2=\dfrac{3+\cos4x}4[/tex]

So the equation reduces to

[tex]\dfrac{3+\cos4x}4=\cos4x[/tex]
[tex]\dfrac34=\dfrac34\cos4x[/tex]
[tex]\cos4x=1[/tex]

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

[tex]4x=2k\pi\implies x=\dfrac{k\pi}2[/tex]

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