You should use the following substitution in the given equation, such that:

`cos theta = t => cos^2 theta = t^2`

`2t^2 - 3t + 1 = 0`

You need to split the midterm `-3t`  into two terms such that:

`2t^2 - 2t - t + 1 = 0`

You need to group the terms such that:

`(2t^2 - 2t) - (t- 1) = 0`

Factoring out 2t in the first group yields:

`2t(t-1) - (t-1) = 0`

You need to factor out `(t-1)`  such that:

`(t-1)(2t-1) = 0`

You need to solve the followings equations such that:

`t - 1 = 0 => t = 1 `

`2t - 1 = 0 => 2t = 1 => t = 1/2`

You need to solve for `theta in [0,2pi)`  the following equations such that:

`cos theta = 1 => theta = 0`

`cos theta = 1/2 => theta = pi/3 and theta = 2pi-pi/3 = 5pi/3`

Hence, evaluating the solutions to the given equation, for `theta in [0,2pi), ` yields `theta = 0, theta = pi/3, theta = 5pi/3.`