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.