We are tasked with solving the trigonometric equation:

$-2\sin^2\theta + \cos\theta + 1 = 0 \quad \text{for} \quad 0 \leq \theta < 2\pi.$

Step 1: Use the Pythagorean identity

Since $\sin^2\theta = 1 - \cos^2\theta$, substitute $\sin^2\theta$ into the equation: $-2(1 - \cos^2\theta) + \cos\theta + 1 = 0.$

Simplify: $-2 + 2\cos^2\theta + \cos\theta + 1 = 0,$ $2\cos^2\theta + \cos\theta - 1 = 0.$

Step 2: Solve the quadratic equation

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

Solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 2$, $b = 1$, and $c = -1$.

$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)},$ $x = \frac{-1 \pm \sqrt{1 + 8}}{4},$ $x = \frac{-1 \pm \sqrt{9}}{4},$ $x = \frac{-1 \pm 3}{4}.$

Thus: $x = \frac{2}{4} = \frac{1}{2}, \quad x = \frac{-4}{4} = -1.$

Step 3: Solve for $\theta$

1. For $x = \cos\theta = \frac{1}{2}$: $\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3} \text{ or } \theta = 5\frac{\pi}{3}.$

2. For $x = \cos\theta = -1$: $\cos\theta = -1 \implies \theta = \pi.$

Final Answer:

The solutions are: $\theta = 0, \frac{\pi}{3}, \pi, 5\frac{\pi}{3}.$

The correct option is: $0, \frac{\pi}{3}, \pi, 5\frac{\pi}{3}$.

Would you like a detailed explanation of any step?

Here are 5 questions to deepen your understanding:

1. What is the importance of the Pythagorean identity in trigonometric equations?
2. How does substituting $\sin^2\theta = 1 - \cos^2\theta$ simplify this equation?
3. What methods can we use to solve quadratic equations involving trigonometric functions?
4. Why are there two solutions for $\cos\theta = \frac{1}{2}$ within the interval $0 \leq \theta < 2\pi$?
5. What does the interval $0 \leq \theta < 2\pi$ represent geometrically?

Tip: Always verify the solutions by substituting back into the original equation to ensure correctness.