Solve the equation \(2 \cos^2 x - 3 \cos x - 2 = 0\) for \(0 \leq x \leq 2\pi\)

Step 1 ：Rewrite the equation as a quadratic equation in terms of \(\cos x\): \(2 \cos^2 x - 3 \cos x - 2 = 0\)

Step 2 ：Factor the quadratic equation: \((2\cos x + 1)(\cos x - 2) = 0\)

Step 3 ：Set each factor equal to zero and solve for \(\cos x\): \(2\cos x + 1 = 0\) and \(\cos x - 2 = 0\)

Step 4 ：Solving \(2\cos x + 1 = 0\) gives \(\cos x = -\frac{1}{2}\). The solutions to this equation in the interval \([0, 2\pi]\) are \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\)

Step 5 ：Solving \(\cos x - 2 = 0\) gives \(\cos x = 2\). However, the range of \(\cos x\) is \([-1, 1]\), so this equation has no solutions.