1. Given $\sin x = \frac{\sqrt{2}}{2}$ and $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, recognize that $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. Since $\frac{\pi}{4}$ is within the interval, $x = \frac{\pi}{4}$. 2. Given $\sin x = -\frac{\sqrt{3}}{2}$ and $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, recall that $\sin \left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$. Since $-\frac{\pi}{3}$ is in the interval, $x = -\frac{\pi}{3}$. 3. Given $\sin x = -\frac{1}{2}$ and $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, we know $\sin \left(-\frac{\pi}{6}\right) = -\frac{1}{2}$. Therefore, $x = -\frac{\pi}{6}$. 4. Given $\sin x = 1$ and $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, recall that $\sin \frac{\pi}{2} = 1$. Hence, $x = \frac{\pi}{2}$. 5. For $\cos x = \frac{1}{2}$ and $x \in [0, \pi]$, since $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\frac{\pi}{3}$ lies in the interval, $x= \frac{\pi}{3}$. 6. For $\cos x = -\frac{\sqrt{3}}{2}$ and $x \in [0, \pi]$, we know $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$. The value $\frac{5\pi}{6}$ lies in the interval, so $x = \frac{5\pi}{6}$. 7. For $\cos x = 0$ and $x \in [0, \pi]$, note that $\cos \frac{\pi}{2} = 0$ and $\frac{\pi}{2}$ is within the interval. Thus, $x= \frac{\pi}{2}$. 8. For $\tan x = -1$ and $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\tan\left(-\frac{\pi}{4}\right) = -1$. Since this is in the interval, $x = -\frac{\pi}{4}$. Final Answers: 1) $\frac{\pi}{4}$ 2) $-\frac{\pi}{3}$ 3) $-\frac{\pi}{6}$ 4) $\frac{\pi}{2}$ 5) $\frac{\pi}{3}$ 6) $\frac{5\pi}{6}$ 7) $\frac{\pi}{2}$ 8) $-\frac{\pi}{4}$