1. **State the problem:** Find all $x$-intercepts of the function $$f(x) = 6 \sin^2 x + 3 \sin x - 3 = 0$$ on the interval $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$. 2. **Rewrite the equation:** Let $$u = \sin x$$. Then the equation becomes a quadratic in $u$: $$6u^2 + 3u - 3 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula: $$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=6$, $b=3$, and $c=-3$. Calculate the discriminant: $$\Delta = 3^2 - 4 \times 6 \times (-3) = 9 + 72 = 81$$ So, $$u = \frac{-3 \pm \sqrt{81}}{12} = \frac{-3 \pm 9}{12}$$ 4. **Find the roots:** - For the plus sign: $$u_1 = \frac{-3 + 9}{12} = \frac{6}{12} = \frac{1}{2}$$ - For the minus sign: $$u_2 = \frac{-3 - 9}{12} = \frac{-12}{12} = -1$$ 5. **Check domain restrictions:** Since $u = \sin x$, and $\sin x$ ranges from $-1$ to $1$, both values are valid. 6. **Find $x$ values for each root on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$:** - For $\sin x = \frac{1}{2}$, the solutions in the interval are: $$x = \frac{\pi}{6}$$ - For $\sin x = -1$, the solution in the interval is: $$x = -\frac{\pi}{2}$$ 7. **Final answer:** The $x$-intercepts of $f(x)$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ are: $$x = -\frac{\pi}{2}, \quad \frac{\pi}{6}$$