To solve the equation \( \cos(2x) - \cos(x) + 1 = 0 \), we can use the double angle identity for cosine, which states that \( \cos(2x) = 2\cos^2(x) - 1 \). 

Substituting this into the equation gives:

\[
2\cos^2(x) - 1 - \cos(x) + 1 = 0
\]

This simplifies to:

\[
2\cos^2(x) - \cos(x) = 0
\]

Factoring out \( \cos(x) \):

\[
\cos(x)(2\cos(x) - 1) = 0
\]

This results in two equations:

1. \( \cos(x) = 0 \)
2. \( 2\cos(x) - 1 = 0 \) or \( \cos(x) = \frac{1}{2} \)

### Solving Each Case:

1. **For \( \cos(x) = 0 \)**:
   - This occurs at \( x = \frac{\pi}{2} + k\pi \), where \( k \) is any integer.

2. **For \( \cos(x) = \frac{1}{2} \)**:
   - This occurs at \( x = \frac{\pi}{3} + 2k\pi \) and \( x = \frac{5\pi}{3} + 2k\pi \), where \( k \) is any integer.

### Final Solutions:

The solutions to the equation \( \cos(2x) - \cos(x) + 1 = 0 \) are:

- \( x = \frac{\pi}{2} + k\pi \)
- \( x = \frac{\pi}{3} + 2k\pi \)
- \( x = \frac{5\pi}{3} + 2k\pi \)

If you have any questions or need further details, feel free to ask!

### Related Questions:
1. What are the graphs of \( \cos(x) \) and \( \cos(2x) \) over one period?
2. How do the solutions change if we consider a specific interval, like \( [0, 2\pi] \)?
3. Can we derive any other identities using the equation?
4. What other trigonometric equations can be solved similarly?
5. How does the cosine function behave at critical points like \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \)?

**Tip:** When solving trigonometric equations, always consider the periodic nature of trigonometric functions to find all possible solutions.

Explore the solution to the trigonometric equation cos(2x) - cos(x) + 1 = 0 using double angle identities. This problem covers key trigonometric concepts and provides detailed solutions for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation