Question

Cos4x-cos2x=0?

Answer 1

\[\cos^4(x)-\cos^2(x)=0\] or
\[\cos(4x)-\cos(2x)=0\]

Answer 2

Second one, no exponents

Answer 3

rewrite \(\cos(4x)\) in terms of sine and cosine of \(2x\)

Answer 4

\[\cos(4x)=\cos^2(2x)-\sin^2(2x)\]

Answer 5

i think that might work or maybe better as
\[2\cos^2(2x)-1\]

Answer 6

then you get
\[2\cos^2(2x)-1-\cos(2x)=0\] or
\[2\cos^2(2x)-\cos(2x)-1=0\] which if you are lucky will factor

Answer 7

hmm somehow this turns in to \[-2 \sin^2(x) (2 \cos(2 x)+1) = 0\] but i am not sure how

Answer 8

this is way too confusing 😧