We are tasked with solving the trigonometric equation:

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

### Step-by-step solution:

1. **Substitute** \( \sin^2(x) \) using the Pythagorean identity:
   \[
   \sin^2(x) = 1 - \cos^2(x)
   \]
   Substituting this into the original equation:

   \[
   4(1 - \cos^2(x)) - \cos(x) - 1 = 0
   \]

   Simplify:

   \[
   4 - 4\cos^2(x) - \cos(x) - 1 = 0
   \]

2. **Simplify** the expression:
   \[
   3 - 4\cos^2(x) - \cos(x) = 0
   \]

   Rearrange it:
   \[
   4\cos^2(x) + \cos(x) - 3 = 0
   \]

3. **Solve** this quadratic equation in terms of \( \cos(x) \). Let \( u = \cos(x) \), so we now have:
   \[
   4u^2 + u - 3 = 0
   \]

4. **Factor** the quadratic equation:
   \[
   (4u - 3)(u + 1) = 0
   \]

   So, the solutions are:
   \[
   u = \frac{3}{4} \quad \text{or} \quad u = -1
   \]

   Recall \( u = \cos(x) \), so:
   \[
   \cos(x) = \frac{3}{4} \quad \text{or} \quad \cos(x) = -1
   \]

5. **Find the general solutions** for \( x \):
   - For \( \cos(x) = \frac{3}{4} \):
     \[
     x = \pm \cos^{-1}\left(\frac{3}{4}\right) + 2n\pi \quad \text{for} \, n \in \mathbb{Z}
     \]

   - For \( \cos(x) = -1 \):
     \[
     x = \pi + 2n\pi \quad \text{for} \, n \in \mathbb{Z}
     \]

Thus, the complete solution is:

\[
x = \pm \cos^{-1}\left(\frac{3}{4}\right) + 2n\pi \quad \text{or} \quad x = \pi + 2n\pi \quad \text{where} \, n \in \mathbb{Z}
\]

Would you like more details or clarifications?

Here are some related questions to deepen understanding:
1. How do you derive the cosine inverse function?
2. What is the significance of the general solution \( x = \pi + 2n\pi \)?
3. How can you apply the quadratic formula to solve trigonometric equations?
4. What are the typical intervals to solve trigonometric equations?
5. How do the Pythagorean identities relate to solving trigonometric equations?

**Tip:** Always check if trigonometric identities like the Pythagorean identity can simplify your equation before attempting to solve.

Explore how to solve the trigonometric equation 4sin^2(x) - cos(x) - 1 = 0 using Pythagorean identities, quadratic factorization, and inverse trigonometric functions. This step-by-step solution is ideal for students in Grades 10-12.

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