3 years ago

Help me please...!!!!!!

Mathematics

2 answers:

goblinko [34]3 years ago

4 0

B would be the answer for sure

riadik2000 [5.3K]3 years ago

4 0

B is the answer i agree with her

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

ICE Princess25 [194]

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

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

$\cos^22x=\dfrac{1+\cos4x}2$
$\implies\sin^4x+\cos^4x=\dfrac{1+\frac{1+\cos4x}2}2=\dfrac{3+\cos4x}4$

So the equation reduces to

$\dfrac{3+\cos4x}4=\cos4x$
$\dfrac34=\dfrac34\cos4x$
$\cos4x=1$

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

$4x=2k\pi\implies x=\dfrac{k\pi}2$