Find all solutions in the interval ​[0,2pi​).
cosine squared x minus 2 cosine x minus 1 equals 0
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
xequals
  
enter your response here
​(Type your answer in radians. Round to four decimal places as needed. Use a comma to separate answers as​ needed.)
B.
There is no solution.

To solve for \( x \) in the interval \( [0, 2\pi) \) for the equation:

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

we can solve this by using a substitution.

1. Let \( u = \cos(x) \). Then \( \cos^2(x) = u^2 \), and the equation becomes:
   \[
   u^2 - 2u - 1 = 0
   \]

2. Now, solve this quadratic equation in terms of \( u \) by using the quadratic formula:
   \[
   u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}
   \]
   Simplifying this:
   \[
   u = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}
   \]

3. This gives:
   \[
   u = 1 + \sqrt{2} \quad \text{and} \quad u = 1 - \sqrt{2}
   \]

4. Since \( u = \cos(x) \), we need \( -1 \leq u \leq 1 \). Checking each value:
   - \( 1 + \sqrt{2} \approx 2.414 \), which is outside the range, so it is invalid.
   - \( 1 - \sqrt{2} \approx -0.414 \), which is within the range.

5. Therefore, we only have \( \cos(x) = 1 - \sqrt{2} \).

6. Now, solve for \( x \):
   \[
   x = \cos^{-1}(1 - \sqrt{2})
   \]
   This yields two solutions in the interval \( [0, 2\pi) \):
   \[
   x \approx 1.9897 \, \text{and} \, x \approx 4.2935
   \]

### Solution
The correct answer is:
\[
x \approx 1.9897, 4.2935
\]

Would you like more details on any part of this solution?

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Here are some related questions to expand on this topic:

1. What are other methods to solve trigonometric equations besides substitution?
2. How do you verify solutions in trigonometric equations?
3. Can trigonometric equations have multiple solutions over different intervals?
4. How can you use identities to simplify trigonometric equations before solving?
5. What are the real-world applications of solving trigonometric equations?

**Tip:** For equations involving \( \cos^2(x) \) or \( \sin^2(x) \), consider using trigonometric identities like \( \cos^2(x) = 1 - \sin^2(x) \) to simplify the problem.

Find all solutions in the interval [0, 2pi) for cos^2(x) - 2cos(x) - 1 = 0. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Explore how to solve the trigonometric equation cos^2(x) - 2cos(x) - 1 = 0 within the interval [0, 2pi). This problem involves substitution, quadratic equations, and the use of the inverse cosine function, making it suitable for students in Grades 10-12.

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