1. **Problem a:** Solve $2 \sin(x) - 1 = 0$ for all $x$. 2. Add 1 to both sides: $$2 \sin(x) = 1$$ 3. Divide both sides by 2: $$\sin(x) = \frac{1}{2}$$ 4. On the unit circle, $\sin(x) = \frac{1}{2}$ at angles $x = \frac{\pi}{6} + 2k\pi$ and $x = \frac{5\pi}{6} + 2k\pi$, where $k$ is any integer. --- 1. **Problem b:** Solve $2 + 2 \cos(x) = 0$ for all $x$. 2. Subtract 2 from both sides: $$2 \cos(x) = -2$$ 3. Divide both sides by 2: $$\cos(x) = -1$$ 4. On the unit circle, $\cos(x) = -1$ at angle $x = \pi + 2k\pi$, where $k$ is any integer. --- 1. **Problem c:** Solve $\sqrt{2} - 2 \sin(x) = 0$ for all $x$. 2. Subtract $\sqrt{2}$ from both sides: $$-2 \sin(x) = -\sqrt{2}$$ 3. Divide both sides by $-2$: $$\sin(x) = \frac{\cancel{-\sqrt{2}}}{\cancel{-2}} = \frac{\sqrt{2}}{2}$$ 4. On the unit circle, $\sin(x) = \frac{\sqrt{2}}{2}$ at angles $x = \frac{\pi}{4} + 2k\pi$ and $x = \frac{3\pi}{4} + 2k\pi$, where $k$ is any integer. --- 1. **Problem d:** Solve $\cos(x) + 3 = 0$ for all $x$. 2. Subtract 3 from both sides: $$\cos(x) = -3$$ 3. Since $\cos(x)$ ranges only between $-1$ and $1$, there is no solution for $\cos(x) = -3$. --- **Summary:** - a) $x = \frac{\pi}{6} + 2k\pi$, $x = \frac{5\pi}{6} + 2k\pi$ - b) $x = \pi + 2k\pi$ - c) $x = \frac{\pi}{4} + 2k\pi$, $x = \frac{3\pi}{4} + 2k\pi$ - d) No solution **Unit circle diagrams** would show points at these angles with coordinates labeled accordingly.