Question

How do you find exact solutions of `cos2x - cosx = 0` in the interval [0,2π) ?

Answer 1

`x=0, (2pi)/3, (4pi)/3`

Explanation:

Recall that

`cos(2x)=cos^2x-sin^2x`.

Now, we have

`cos^2x-sin^2x-cosx=0`

However, we want our equation in terms of only one trigonometric function. We can easily get everything in terms of cosine:

`sin^2x+cos^2x=1`

`sin^2x=1-cos^2x`

Thus,

`cos^2x-(1-cos^2x)-cosx=0`

`2cos^2x-cosx-1=0`

This resembles a quadratic function, except it is composed of cosines. So, let's factor it as we would any quadratic function. I'll leave that part up to you, as it's just a matter of factoring as you would factor any other quadratic however you like:

`(cosx-1)(2cosx+1)=0`

Now, we solve the following:

`cosx-1=0`
`2cos+1=0`

`cosx=1`

Implies that `x=0` since we're working in `[0, 2pi).` It's the only value yielding one for cosine.

`2cosx=-1`
`cosx=-1/2`

Implies that `x=(2pi)/3, (4pi)/3` as these two values return `-1/2` for cosine in the interval `[0, 2pi)`