Let's solve the given problem step-by-step.

(a) Given the equation:

$2 \sin^2 x - 3 \sin x = 2$

We are to find the exact values of $\sin x$.

1. Let $y = \sin x$, then the equation becomes: $2y^2 - 3y = 2$

2. Rearrange it to form a standard quadratic equation: $2y^2 - 3y - 2 = 0$

3. Use the quadratic formula to solve for $y$: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   Here, $a = 2$, $b = -3$, and $c = -2$.

4. Substituting these values: $y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}$ $y = \frac{3 \pm \sqrt{9 + 16}}{4}$ $y = \frac{3 \pm \sqrt{25}}{4}$ $y = \frac{3 \pm 5}{4}$

5. This gives two solutions: $y = \frac{8}{4} = 2 \quad \text{or} \quad y = \frac{-2}{4} = -0.5$

Since $\sin x$ must be between -1 and 1, the valid solution is: $\sin x = -0.5$

(b) Solve the equation $2 \sin^2 x - 3 \sin x = 2$ for $0^\circ < x < 360^\circ$:

From part (a), we found: $\sin x = -0.5$

The solutions for $\sin x = -0.5$ in the interval $0^\circ < x < 360^\circ$ are: $x = 210^\circ \quad \text{or} \quad x = 330^\circ$

Final Answer:

• (a) The exact value of $\sin x$ is $-0.5$.
• (b) The solutions for $x$ in the interval $0^\circ < x < 360^\circ$ are: $x = 210^\circ \quad \text{or} \quad x = 330^\circ$

Would you like a detailed breakdown of any specific step?

Related Questions:

1. How do you derive the quadratic formula from a standard quadratic equation?
2. What are the typical steps to solve a trigonometric equation?
3. How do you determine the angles that satisfy a trigonometric equation within a specific interval?
4. Why is it necessary to check if solutions lie within the valid range for trigonometric functions?
5. How can you solve similar trigonometric equations involving cosine or tangent?

Tip: Always verify that solutions fall within the specified interval and range for trigonometric functions. This ensures accurate answers!