To solve the equation cos^3 x - 3cos^2 x + 2 cos x = 0, we can factor out a cos x:

cos x (cos^2 x - 3cos x + 2) = 0

Now we have a quadratic equation in terms of cos x:

cos x (cos x - 1)(cos x - 2) = 0

Setting each factor to zero:

cos x = 0
cos x = 1
cos x = 2

However, the range of cosine function is from -1 to 1, so the value of cos x cannot be 2. Therefore, we only consider the first two equations:

cos x = 0 --> x = π/2 + nπ, where n is an integer
cos x = 1 --> x = 2nπ, where n is an integer

So the solutions to the equation cos^3 x - 3cos^2 x + 2 cos x = 0 are x = π/2 + nπ and x = 2nπ, where n is an integer.