cos2x = cosx solve for interval (0 2π)

Question

cos2x = cosx solve for interval (0 2π)

algebraic means without calculator 

Answer 1

cos(2x) = cosx

 

2cos²x - 1 = cosx

 

2cos²x - cosx - 1 = 0

 

Let u = cosx.

 

Then 2u² - u - 1 = 0

 

      (2u+1)(u-1) = 0

 

         u = -1/2 or u = 1

 

So, cosx = -1/2 or cosx = 1

 

If cosx = 1, then x = 0, ±2π, ±4π, ...

 

If cosx = -1/2, then x = 2π/3+2kπ, 4π/3+2kπ, k = 0, ±1, ±2, ...

 

Solutions in the interval (0, 2π):  2π/3, 4π/3

 

Answer 2

move aside so that cos2x - cosx = 0 

integrated each other so it becomes :

1/2sin2x - sinx

for interval (0,2pi) so :

(1/2sin0 - sin0) - (1/2sin4pi - sin2pi)

sin 0 = 0

sin 2pi = sin 360 = 0

sin 4pi = sin 720 = 0

so the ansewe is 0