1. **Problem statement:** Find all values of $t \in [0, 2\pi]$ that satisfy the following equations: (a) $\sin(t) = -\frac{1}{2}$ (b) $\cos(t) = \frac{1}{2}$ (c) $\tan(t) = \frac{1}{\sqrt{3}}$ 2. **Formulas and important rules:** - For sine and cosine, solutions in $[0, 2\pi]$ can be found using the unit circle and symmetry. - For tangent, solutions repeat every $\pi$. 3. **Solve (a) $\sin(t) = -\frac{1}{2}$:** - The reference angle where $\sin(\theta) = \frac{1}{2}$ is $\theta = \frac{\pi}{6}$. - Since sine is negative in the third and fourth quadrants, solutions are: $$t = \pi + \frac{\pi}{6} = \frac{7\pi}{6}, \quad t = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$$ 4. **Solve (b) $\cos(t) = \frac{1}{2}$:** - The reference angle where $\cos(\theta) = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$. - Cosine is positive in the first and fourth quadrants, so solutions are: $$t = \frac{\pi}{3}, \quad t = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$ 5. **Solve (c) $\tan(t) = \frac{1}{\sqrt{3}}$:** - The reference angle where $\tan(\theta) = \frac{1}{\sqrt{3}}$ is $\theta = \frac{\pi}{6}$. - Tangent is positive in the first and third quadrants, so solutions are: $$t = \frac{\pi}{6}, \quad t = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$ **Final answers:** (a) $t = \frac{7\pi}{6}, \frac{11\pi}{6}$ (b) $t = \frac{\pi}{3}, \frac{5\pi}{3}$ (c) $t = \frac{\pi}{6}, \frac{7\pi}{6}$