Question

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

cos(4x) - cos(2x) = 0

Answer 1

is that an exponent or no?

Answer 2

no

Answer 3

@Luigi0210 @dumbcow can u help me

Answer 4

use double angle formula
\[\cos 2 \theta = 2\cos^{2} \theta -1\]
\[\rightarrow (2\cos^{2} (2x) -1)-\cos(2x) = 0\]
treat it like a quadratic equation and solve for "cos 2x "

Answer 5

How do u write it as a quad

Answer 6

\[2u^{2} -u-1 = 0\]
where u=cos 2x
this will factor

Answer 7

how do u get cos2x = u

Answer 8

magic ...haha no i made it up to help you see the quadratic form
its called a substitution

Answer 9

so i dont need the cos2x

Answer 10

am i just subbing it because i can

Answer 11

exactly, it helps to factor it ... but in the end you will have ...cos 2x = ?

Answer 12

now i have
2x=0 2x=cos of (-1/2)

Answer 13

@dumbcow

Answer 14

@torobi From where @dumbcow left off, after you have factored the quadratic where you made the substitution u = cos(2x), you get:\[\bf 2u^2-u-1=0 \implies (2u+1)(u-1)=0 \implies u = -\frac{1}{2} \ or \ u=1\]After we have factored and achieved our solution, we bring cos(2x) back in and replace u with cos(2x) again:\[\bf \cos(2x)=u \implies \cos(2x)=-\frac{1}{2}\]\[\bf \cos(2x)=u \implies \cos(2x)=1\]Now we solve for 'x' in each case and get our solutions to whatever 'x' is. Can you do that?

@torobi