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The given equation is:

\[
12 \cos^2(x) + \cos(x) - 13 = 0
\]

This is a quadratic equation in terms of \(\cos(x)\), and we can solve it by factoring or using the quadratic formula.

### Step 1: Set up the equation for factoring
We rewrite it as:

\[
12y^2 + y - 13 = 0
\]

where \(y = \cos(x)\).

### Step 2: Factor the quadratic equation
We look for factors of the quadratic equation:

\[
12y^2 + y - 13 = (3y - 1)(4y + 13) = 0
\]

### Step 3: Solve each factor
Set each factor equal to zero:

1. \(3y - 1 = 0\) gives \(y = \frac{1}{3}\)
2. \(4y + 13 = 0\) gives \(y = -\frac{13}{4}\)

### Step 4: Return to \(\cos(x)\)
Since \(y = \cos(x)\), we have:

1. \(\cos(x) = \frac{1}{3}\)
2. \(\cos(x) = -\frac{13}{4}\) (This is not valid, since \(\cos(x)\) must lie between -1 and 1)

### Step 5: Solve for \(x\)
For \(\cos(x) = \frac{1}{3}\), we take the inverse cosine:

\[
x = \cos^{-1}\left(\frac{1}{3}\right)
\]

Thus, the solution is:

\[
x = \cos^{-1}\left(\frac{1}{3}\right)
\]

If you'd like more details on this or have questions, feel free to ask!

### Related Questions:
1. What is the general solution for trigonometric equations involving cosine?
2. How does factoring work for quadratic equations?
3. What are the domain and range of \(\cos(x)\)?
4. Can you explain the inverse trigonometric functions in detail?
5. How do we verify if a trigonometric solution is valid in the context of the problem?

**Tip**: Always check the domain of trigonometric functions, as not all values for cosine or sine may fall within the valid range of \([-1, 1]\).

12 cos^2(x) + cos(x) - 13 = 0 (Factor and simplify)

Learn how to factor and simplify the trigonometric equation 12 cos^2(x) + cos(x) - 13 = 0. This step-by-step guide explores solving quadratic equations with trigonometric functions, including factoring techniques and properties of the cosine function. Suitable for students in Grades 9-12.

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