To solve the equation $\cos 2x - \cos x + 1 = 0$, we can use trigonometric identities and substitution.

Step 1: Apply the double-angle identity for cosine

Recall that: $\cos 2x = 2\cos^2 x - 1$ Substitute this into the equation: $2\cos^2 x - 1 - \cos x + 1 = 0$

Step 2: Simplify the equation

Combine like terms: $2\cos^2 x - \cos x = 0$

Step 3: Factor the equation

Factor out $\cos x$: $\cos x (2\cos x - 1) = 0$

This gives us two possible cases:

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

Step 4: Solve each case

Case 1: $\cos x = 0$

If $\cos x = 0$, then: $x = \frac{\pi}{2} + n\pi \quad \text{for integer } n$

Case 2: $2\cos x - 1 = 0$

Solve for $\cos x$: $2\cos x = 1$ $\cos x = \frac{1}{2}$ This gives: $x = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2n\pi \quad \text{for integer } n$

Final Solution

The solutions to the equation $\cos 2x - \cos x + 1 = 0$ are: $x = \frac{\pi}{2} + n\pi, \quad x = \frac{\pi}{3} + 2n\pi, \quad \text{and} \quad x = \frac{5\pi}{3} + 2n\pi$ for integer $n$.

Would you like further details on any step, or have any questions?

Here are some related questions for further exploration:

1. How would this solution change if the equation were $\cos 2x - 2\cos x + 1 = 0$?
2. What are the general solutions for $\cos^2 x - \cos x = 0$?
3. How can we derive the double-angle identities from first principles?
4. How would you solve $\sin 2x - \sin x = 0$ using similar techniques?
5. Can you solve $\cos 3x - \cos x = 0$ using trigonometric identities?

Tip: When solving trigonometric equations, always check if identities can simplify expressions before expanding further.