Let's solve the equation step-by-step:

### Given equation:
\[
7\cos^2x - 3\sin^2x = -1 - \cos x
\]

### Step 1: Use the Pythagorean identity
Recall the identity:
\[
\sin^2x = 1 - \cos^2x
\]

Substitute this into the equation:
\[
7\cos^2x - 3(1 - \cos^2x) = -1 - \cos x
\]

### Step 2: Simplify the equation
Expand the equation:
\[
7\cos^2x - 3 + 3\cos^2x = -1 - \cos x
\]

Combine like terms:
\[
(7\cos^2x + 3\cos^2x) - 3 = -1 - \cos x
\]
\[
10\cos^2x - 3 = -1 - \cos x
\]

### Step 3: Move all terms to one side
Add \(1\) and \(\cos x\) to both sides:
\[
10\cos^2x + \cos x - 2 = 0
\]

### Step 4: Solve the quadratic equation
This is a quadratic equation in terms of \(\cos x\):
\[
10u^2 + u - 2 = 0
\]
where \(u = \cos x\).

To solve this quadratic equation, use the quadratic formula:
\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 10\), \(b = 1\), and \(c = -2\):
\[
u = \frac{-1 \pm \sqrt{1^2 - 4 \times 10 \times (-2)}}{2 \times 10}
\]
\[
u = \frac{-1 \pm \sqrt{1 + 80}}{20}
\]
\[
u = \frac{-1 \pm \sqrt{81}}{20}
\]
\[
u = \frac{-1 \pm 9}{20}
\]

### Step 5: Find the roots
\[
u_1 = \frac{-1 + 9}{20} = \frac{8}{20} = \frac{2}{5}
\]

\[
u_2 = \frac{-1 - 9}{20} = \frac{-10}{20} = -\frac{1}{2}
\]

### Step 6: Find \(x\)
1. For \(u_1 = \frac{2}{5}\):
   \[
   \cos x = \frac{2}{5}
   \]
   \[
   x = \arccos\left(\frac{2}{5}\right)
   \]
   The solutions for \(x\) are:
   \[
   x = \arccos\left(\frac{2}{5}\right) \quad \text{and} \quad x = 2\pi - \arccos\left(\frac{2}{5}\right)
   \]

2. For \(u_2 = -\frac{1}{2}\):
   \[
   \cos x = -\frac{1}{2}
   \]
   \[
   x = \arccos\left(-\frac{1}{2}\right)
   \]
   The solutions for \(x\) are:
   \[
   x = \frac{2\pi}{3} \quad \text{and} \quad x = \frac{4\pi}{3}
   \]

### Final solutions
The general solutions for \(x\) are:
\[
x = \arccos\left(\frac{2}{5}\right) + 2n\pi \quad \text{or} \quad x = 2\pi - \arccos\left(\frac{2}{5}\right) + 2n\pi
\]
\[
x = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2n\pi
\]

### Summary
The equation \(7\cos^2x - 3\sin^2x = -1 - \cos x\) has the solutions \(x = \arccos\left(\frac{2}{5}\right)\), \(x = 2\pi - \arccos\left(\frac{2}{5}\right)\), \(x = \frac{2\pi}{3}\), and \(x = \frac{4\pi}{3}\), along with their periodic counterparts.

Would you like more details, or do you have any questions?

### 5 Follow-up Questions
1. How do you solve a quadratic equation in general?
2. What is the significance of the Pythagorean identity in trigonometry?
3. How do you find the general solution for trigonometric equations?
4. Why does the quadratic formula work for any quadratic equation?
5. How can we verify the solutions to a trigonometric equation?

### Tip
Always double-check the domain and range of trigonometric functions when solving equations to ensure all potential solutions are considered.

Explore the detailed solution to the trigonometric equation 7cos^2x - 3sin^2x = -1 - cosx. This solution covers key concepts in trigonometry, such as the Pythagorean identity, solving quadratic equations in cosine, and utilizing the quadratic formula. Ideal for high school students studying trigonometry.

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