Question

Find all solutions to the equation in the interval [0, 2π).
cos 2x - cos x = 0

Answer 1

Have you any formulas for \(\cos(2x)\) in terms of \(\cos(x)\)?

Answer 2

change cos 2x to 2cos^2 x - 1

Answer 3

Look at your trig identities

Answer 4

cosx=1/secx

Answer 5

i am looking. i understand that. i dont understand the solving part...

Answer 6

?? slaaibak already gave to you the one that will help.

Answer 7

but why. i wanna understand.

Answer 8

cosa = cos b

why don;t you generalize a simply ? In terms of b

Answer 9

because, when you only have a cos x 's, you can factorise it, which makes solving it easier

Answer 10

because i'm not a genius. this is hard to teach yourself

Answer 11

so now i have 2cos^(2)x-1-cosx = 0?

Answer 12

If it were this, \(2x^{2} - x - 1 = 0\), what would you do to solve it?

Answer 13

-b (plus or minus) the square root of b^2-4(a)(c)/2a

Answer 14

Perfect. Do that with "cos(x)" instead.

Answer 15

a=2cos^2x
b=-1
c= nothing?

Answer 16

No.

\(2\cos^{2}(x) - 1 - \cos(x) = 0\)

or, in standard form,

\(2\cos^{2}(x) - \cos(x) - 1 = 0\)

a = 2
b = -1
c = -1

Now write the answer:

\(\cos(x) = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

Do it EXACTLY like you would with \(2x^{2} - x - 1 = 0\)

It is the same if you have \(2(Frog)^{2} - (Frog) - 1 = 0\). Write \(Frog = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

It is the same if you have \(2(y^{2})^{2} - (y^{2}) - 1 = 0\). Write \(y^{2} = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

Answer 17

\[\frac{ -cosx \pm \sqrt{cosx^2-4*2\cos^2x} }{ 2*2\cos^2x }\]

Answer 18

Why did you write that?

Please go read that last response again. It starts out "cos(x) =" and cos(x) never appears again.

Answer 19

cos(2x) - cos(x) = 0
==> [2cos^2(x) - 1] - cos(x) = 0
==> 2cos^2(x) - cos(x) - 1 = 0
==> [2cos(x) + 1][cos(x) - 1] = 0
==> cos(x) = -1/2 and cos(x) = 1.

Answer 20

x = 0, 2π/3, 4π/3.

Answer 21

Absolutely, factoring is preferred to the Quadratic Formula.