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The given equation is:

\[
\cos^2 x + 6 \cos x + 4 = 0
\]

This is a quadratic equation in terms of \( \cos x \). To solve it, we can treat \( \cos x \) as a variable (say, \( u \)) and rewrite the equation:

\[
u^2 + 6u + 4 = 0 \quad \text{where} \quad u = \cos x
\]

Now, apply the quadratic formula:

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( u^2 + 6u + 4 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = 6 \)
- \( c = 4 \)

Substitute these values into the quadratic formula:

\[
u = \frac{-6 \pm \sqrt{6^2 - 4(1)(4)}}{2(1)}
\]

\[
u = \frac{-6 \pm \sqrt{36 - 16}}{2}
\]

\[
u = \frac{-6 \pm \sqrt{20}}{2}
\]

\[
u = \frac{-6 \pm 2\sqrt{5}}{2}
\]

\[
u = -3 \pm \sqrt{5}
\]

Thus, the two possible solutions for \( u \) are:

\[
u_1 = -3 + \sqrt{5} \quad \text{and} \quad u_2 = -3 - \sqrt{5}
\]

Now, evaluate the numerical values of \( u_1 \) and \( u_2 \):

\[
u_1 = -3 + \sqrt{5} \approx -0.7639
\]
\[
u_2 = -3 - \sqrt{5} \approx -5.2361
\]

Since \( \cos x \) must be in the range \([-1, 1]\), only \( u_1 \approx -0.7639 \) is a valid solution.

### Solving for \( x \):
Now, find \( x \) such that \( \cos x = -0.7639 \). Use the inverse cosine function:

\[
x = \cos^{-1}(-0.7639)
\]

This gives two possible solutions in the interval \([0^\circ, 360^\circ]\):

\[
x_1 \approx \cos^{-1}(-0.7639) \approx 141.57^\circ
\]
\[
x_2 = 360^\circ - x_1 = 360^\circ - 141.57^\circ \approx 218.43^\circ
\]

Thus, the solutions are approximately:

\[
x \approx 141.57^\circ, 218.43^\circ
\]

### Final Answer:
The solutions in degrees are \( 141.57^\circ \) and \( 218.43^\circ \).

Would you like more details or clarifications?

Here are 5 related questions to consider:
1. How does the range of cosine affect the solutions?
2. What is the significance of the quadratic formula in this context?
3. How do we verify the solutions using a calculator or by substitution?
4. What are other methods to solve trigonometric equations like this one?
5. How would the problem change if the interval was in radians instead of degrees?

**Tip:** Always check the validity of the roots for trigonometric equations within the defined range of the trigonometric function.

Solve, finding all solutions in [0, 360°) for the equation cos^2(x) + 6cos(x) + 4 = 0.

This step-by-step guide explains how to solve the trigonometric quadratic equation cos^2(x) + 6cos(x) + 4 = 0, using the quadratic formula and properties of the cosine function. It covers key trigonometry concepts suitable for students in Grades 9-12.

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