He means substituting y = cos(x)
and then solving that equation.
To solve a polynomial, find a solution to that equation (in this case, y = 1 satisfies the equation) Therefore its a solution to the polynomial and (y - 1) is a factor. Divide the polynomial by that factor, and you should get a quadratic. Apply the quadratic formula to that to get the other two answers.
Remember, when you have your solutions, don't forget to change your y back into cos(x)
For a more indepth look at solving polynomails look at this site Solving Polynomial Equations
so it becomes
2y^3 - y^2 - 2y + 1 = 0
solve for y as you would a polynomial then sub back into the y
2 cos theta -1 = 0
cos theta + 1 = 0
cos theta -1 = 0
depends what range the solution is required
between 0 and 2 pi (inclusive)
i get: 0, pi/3, pi, 5pi/3, 2pi