1. **Problem 1: Find all solutions between 0 and 2\pi to the equation $\sin(3x) = -\frac{1}{\sqrt{2}}$.** 2. The general solution for $\sin \theta = k$ is $\theta = \arcsin(k) + 2n\pi$ or $\theta = \pi - \arcsin(k) + 2n\pi$ for any integer $n$. 3. Here, $\theta = 3x$ and $k = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$. 4. Calculate $\arcsin\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}$. 5. So, the solutions for $3x$ are: $$3x = -\frac{\pi}{4} + 2n\pi \quad \text{or} \quad 3x = \pi - \left(-\frac{\pi}{4}\right) + 2n\pi = \pi + \frac{\pi}{4} + 2n\pi = \frac{5\pi}{4} + 2n\pi$$ 6. Simplify and solve for $x$: $$x = \frac{-\frac{\pi}{4} + 2n\pi}{3} = -\frac{\pi}{12} + \frac{2n\pi}{3}$$ $$x = \frac{\frac{5\pi}{4} + 2n\pi}{3} = \frac{5\pi}{12} + \frac{2n\pi}{3}$$ 7. Find all $x$ in $[0, 2\pi)$ by testing integer values of $n$: - For $x = -\frac{\pi}{12} + \frac{2n\pi}{3}$: - $n=1$: $x = -\frac{\pi}{12} + \frac{2\pi}{3} = \frac{7\pi}{12}$ - $n=2$: $x = -\frac{\pi}{12} + \frac{4\pi}{3} = \frac{15\pi}{12} = \frac{5\pi}{4}$ - $n=3$: $x = -\frac{\pi}{12} + 2\pi = \frac{23\pi}{12}$ - For $x = \frac{5\pi}{12} + \frac{2n\pi}{3}$: - $n=0$: $x = \frac{5\pi}{12}$ - $n=1$: $x = \frac{5\pi}{12} + \frac{2\pi}{3} = \frac{13\pi}{12}$ - $n=2$: $x = \frac{5\pi}{12} + \frac{4\pi}{3} = \frac{21\pi}{12} = \frac{7\pi}{4}$ 8. The solutions in $[0, 2\pi)$ are: $$x = \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{13\pi}{12}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{23\pi}{12}$$ --- 9. **Problem 2: Express the complex number $z = 1 - 2i$ in polar form.** 10. The polar form of a complex number $z = a + bi$ is: $$z = r(\cos \theta + i \sin \theta) = r e^{i\theta}$$ where $$r = \sqrt{a^2 + b^2}$$ $$\theta = \arctan\left(\frac{b}{a}\right)$$ 11. Calculate $r$: $$r = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$ 12. Calculate $\theta$: $$\theta = \arctan\left(\frac{-2}{1}\right) = \arctan(-2)$$ Since $a=1 > 0$ and $b=-2 < 0$, $\theta$ is in the fourth quadrant. 13. Approximate $\arctan(-2) \approx -1.107$ radians or equivalently: $$\theta = 2\pi - 1.107 = 5.176 \text{ radians}$$ 14. Therefore, the polar form is: $$z = \sqrt{5} \left( \cos 5.176 + i \sin 5.176 \right) = \sqrt{5} e^{i 5.176}$$