To solve the equation $2\cos^2x + \cos x = 1$ within the interval $-2\pi$ to $2\pi$, follow these steps:

---

Step 1: Rewrite the equation in standard quadratic form

The given equation is: $2\cos^2x + \cos x - 1 = 0$

Let $y = \cos x$. The equation becomes: $2y^2 + y - 1 = 0$

---

Step 2: Solve the quadratic equation

Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 2$, $b = 1$, $c = -1$. Substituting these values: $y = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}$ $y = \frac{-1 \pm \sqrt{1 + 8}}{4}$ $y = \frac{-1 \pm \sqrt{9}}{4}$ $y = \frac{-1 \pm 3}{4}$

Thus: $y = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}, \quad y = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$

---

Step 3: Solve for $\cos x$

$\cos x = \frac{1}{2}, \quad \cos x = -1$

Case 1: $\cos x = \frac{1}{2}$

The solutions occur at: $x = \pm \frac{\pi}{3}, \pm \frac{5\pi}{3}$

Case 2: $\cos x = -1$

The solution occurs at: $x = \pi$

---

Step 4: Combine all solutions within $-2\pi \leq x \leq 2\pi$

The solutions are: $x = -\frac{5\pi}{3}, -\frac{\pi}{3}, \pi, \frac{\pi}{3}, \frac{5\pi}{3}$

---

Final Answer:

$x = -\frac{5\pi}{3}, -\frac{\pi}{3}, \pi, \frac{\pi}{3}, \frac{5\pi}{3}$

Would you like a detailed explanation for any of these steps?

---

Follow-up questions to enhance understanding:

1. How is the quadratic formula applied to trigonometric equations?
2. What are the exact values of cosine at key angles like $\pi/3$ or $\pi$?
3. Why does $\cos x$ repeat its values periodically?
4. How can you verify solutions using a graph of $2\cos^2x + \cos x - 1$?
5. What is the significance of the interval $-2\pi$ to $2\pi$ in solving this problem?

Tip: When solving trigonometric equations, always consider the periodic nature of sine and cosine functions!