Question

Find all solutions to the equation in the interval [0, 2ð).

cos 4x - cos 2x = 0

Answer 1

Let u= 2x
\[\cos(2u)=\cos^2(u)-\sin^2(u)=2\cos^2(u)-1\]

It's not a quadratic.

Answer 2

What do you mean it's not quadratic?

Answer 3

It IS quadratic.

Answer 4

0 pi/3 2pi/3 4pi/3 5pi/3, pi?

Answer 5

\[2\cos^2(u)-\cos(u)-1=0\]
\[(2\cos(u)+1)(\cos(u)-1)=0\]
\[\cos(u)-1=0, \cos(u)=1, u=2x=0,2 \pi\]
or
\[2\cos(u)+1=0, \cos(u)=-0.5, u=2x=5\pi/8, 7 \pi/8\]

Answer 6

I don't understand

Answer 7

What part?

Answer 8

I'm confused by the u

Answer 9

I simply called the quantity '2x' 'u' for simplicity.

Answer 10

I just realized that, I think I got it! Thank you for the help