e general solutions of the followin \( \sqrt{3} \cos \theta-3 \sin \theta=0 \) \( (\cos \theta+\sin \theta)(1-2 \sin \theta)=0 \)

1) Solve \[ \sqrt{3} \cos \theta - 3 \sin \theta = 0. \] **Step 1.** Assume \(\cos\theta \neq 0\) and divide both sides by \(\cos\theta\): \[ \sqrt{3} - 3 \tan\theta = 0. \] **Step 2.** Solve for \(\tan\theta\): \[ 3 \tan\theta = \sqrt{3} \quad \Longrightarrow \quad \tan\theta = \frac{\sqrt{3}}{3}. \] **Step 3.** Recognize that \[ \tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}. \] Thus, the general solution is: \[ \theta = \frac{\pi}{6} + k\pi, \quad k \in \mathbb{Z}. \] --- 2) Solve \[ (\cos \theta + \sin \theta)(1-2 \sin \theta)=0. \] This equation is satisfied when at least one of the factors is zero. **Case 1:** \[ \cos \theta + \sin \theta = 0. \] **Step 1.** Rearrange: \[ \cos \theta = -\sin \theta. \] **Step 2.** Divide both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)): \[ 1 = -\tan\theta \quad \Longrightarrow \quad \tan\theta = -1. \] **Step 3.** The general solution for \(\tan\theta = -1\) is: \[ \theta = -\frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}. \] An equivalent way to represent it is: \[ \theta = \frac{3\pi}{4} + k\pi, \quad k \in \mathbb{Z}. \] **Case 2:** \[ 1-2 \sin \theta = 0. \] **Step 1.** Solve for \(\sin \theta\): \[ 2\sin \theta = 1 \quad \Longrightarrow \quad \sin \theta = \frac{1}{2}. \] **Step 2.** The solutions for \(\sin \theta = \frac{1}{2}\) are: \[ \theta = \frac{\pi}{6}+2k\pi \quad \text{or} \quad \theta = \frac{5\pi}{6}+2k\pi, \quad k \in \mathbb{Z}. \] --- **Final Answers:** 1) For \(\sqrt{3} \cos \theta - 3 \sin \theta = 0\): \[ \theta = \frac{\pi}{6} + k\pi, \quad k \in \mathbb{Z}. \] 2) For \((\cos \theta + \sin \theta)(1-2 \sin \theta)=0\): - Either \[ \theta = -\frac{\pi}{4}+ k\pi \quad (\text{or equivalently } \theta = \frac{3\pi}{4}+ k\pi), \quad k \in \mathbb{Z}, \] - or \[ \theta = \frac{\pi}{6}+ 2k\pi \quad \text{or} \quad \theta = \frac{5\pi}{6}+2k\pi, \quad k \in \mathbb{Z}. \]