We are given the equation:

$2\cos^2x - \cos x = 1$

Let's solve it step by step.

Step 1: Move all terms to one side of the equation

We'll start by moving the 1 from the right side to the left side:

$2\cos^2x - \cos x - 1 = 0$

Step 2: Factor the quadratic equation

We can factor this equation as a quadratic in terms of $\cos x$. Let $y = \cos x$, so the equation becomes:

$2y^2 - y - 1 = 0$

Now, factor the quadratic equation:

$(2y + 1)(y - 1) = 0$

Step 3: Solve for $y$

Set each factor equal to zero:

1. $2y + 1 = 0$

   $y = -\frac{1}{2}$

2. $y - 1 = 0$

   $y = 1$

Step 4: Solve for $x$

Recall that $y = \cos x$, so now we solve:

1. $\cos x = -\frac{1}{2}$

   $x = \cos^{-1}\left(-\frac{1}{2}\right)$

   The solutions within the interval $0 \leq x < 2\pi$ are:

   $x = \frac{2\pi}{3}, \frac{4\pi}{3}$

2. $\cos x = 1$

   $x = \cos^{-1}(1)$

   The solution within the interval $0 \leq x < 2\pi$ is:

   $x = 0$

Final Answer:

The solutions are:

$x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$

Would you like more details on any step or have any questions?

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Here are 5 follow-up questions related to the problem:

1. How can we derive the general solutions for $\cos x = -\frac{1}{2}$ beyond the interval $0 \leq x < 2\pi$?
2. What are the implications of the periodic nature of the cosine function on the solutions?
3. How can we solve similar trigonometric equations involving sine or tangent functions?
4. What are the differences in solving equations involving $\cos^2x$ versus $\sin^2x$?
5. How does the factorization process work for quadratic equations in trigonometric functions?

Tip: Always check the interval or domain restrictions when solving trigonometric equations to ensure you find all relevant solutions.