To solve the equation $2 \sin^2(\theta) + \cos(\theta) = 1$:

Step 1: Rewrite $\sin^2(\theta)$ using the Pythagorean identity

The Pythagorean identity is: $\sin^2(\theta) = 1 - \cos^2(\theta)$ Substitute this into the equation: $2(1 - \cos^2(\theta)) + \cos(\theta) = 1$

Step 2: Simplify the equation

$2 - 2\cos^2(\theta) + \cos(\theta) = 1$ $-2\cos^2(\theta) + \cos(\theta) + 1 = 0$

Step 3: Rearrange into standard quadratic form

$2\cos^2(\theta) - \cos(\theta) - 1 = 0$

Step 4: Solve the quadratic equation

Let $x = \cos(\theta)$, so the equation becomes: $2x^2 - x - 1 = 0$ Solve using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 2$, $b = -1$, $c = -1$. Substituting: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)}$ $x = \frac{1 \pm \sqrt{1 + 8}}{4}$ $x = \frac{1 \pm \sqrt{9}}{4}$ $x = \frac{1 \pm 3}{4}$ $x = \frac{1 + 3}{4} = 1 \quad \text{or} \quad x = \frac{1 - 3}{4} = -\frac{1}{2}$

Step 5: Back-substitute $x = \cos(\theta)$

$\cos(\theta) = 1 \quad \text{or} \quad \cos(\theta) = -\frac{1}{2}$

Step 6: Solve for $\theta$

1. If $\cos(\theta) = 1$: $\theta = 0 + 2k\pi, \quad k \in \mathbb{Z}$

2. If $\cos(\theta) = -\frac{1}{2}$: $\theta = \pi - \frac{\pi}{3} + 2k\pi = \frac{2\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$ or $\theta = \pi + \frac{\pi}{3} + 2k\pi = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$

Final Answer:

$\theta = 0 + 2k\pi, \, \frac{2\pi}{3} + 2k\pi, \, \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$

Let me know if you want further clarification or visual steps.

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Expanding Questions:

1. How does the Pythagorean identity help simplify trigonometric equations?
2. Why is the quadratic formula useful in solving trigonometric equations?
3. What is the general solution for $\cos(\theta) = -\frac{1}{2}$?
4. How can you verify these solutions on a unit circle?
5. What is the significance of the parameter $k$ in the general solution?

Tip: Always remember the Pythagorean identities to rewrite and simplify trigonometric expressions!