The equation provided is:

$2\sin^2\theta + \sin\theta = 1 \quad \text{for } 0^\circ \leq \theta \leq 360^\circ.$

Step 1: Rearrange the equation

Rearrange the equation into a standard quadratic form:

$2\sin^2\theta + \sin\theta - 1 = 0.$

Let $x = \sin\theta$, so the equation becomes:

$2x^2 + x - 1 = 0.$

Step 2: Solve the quadratic equation

We use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$

where $a = 2$, $b = 1$, and $c = -1$. Substituting:

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

Simplify:

$x = \frac{-1 \pm 3}{4}.$

This gives two solutions:

$x = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}, \quad x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1.$

Step 3: Solve for $\theta$

Recall that $x = \sin\theta$, so:

1. $\sin\theta = \frac{1}{2}$
2. $\sin\theta = -1$

Case 1: $\sin\theta = \frac{1}{2}$

From the unit circle, $\sin\theta = \frac{1}{2}$ occurs at:

$\theta = 30^\circ, \quad \theta = 150^\circ.$

Case 2: $\sin\theta = -1$

From the unit circle, $\sin\theta = -1$ occurs at:

$\theta = 270^\circ.$

Step 4: Combine all solutions

The solutions for $\theta$ are:

$\boxed{\theta = 30^\circ, 150^\circ, 270^\circ.}$

Would you like additional clarification or a graph of the solution?

Related Questions:

1. What are the key properties of the sine function that helped solve this problem?
2. How can the quadratic formula be applied to trigonometric equations in general?
3. Why are there exactly three solutions in this case within $0^\circ \leq \theta \leq 360^\circ$?
4. What would change if the range of $\theta$ were different, e.g., $0^\circ \leq \theta \leq 720^\circ$?
5. How can a graph of $y = 2\sin^2\theta + \sin\theta - 1$ help verify these solutions?

Tip:

Always verify solutions to trigonometric equations within the given interval to ensure no solutions are missed.