2) \( \frac{1 \cdot \cos \theta}{\sin }=\frac{\sin \theta}{1+\cos \theta} \) 3) \( \frac{1+2 \sin \theta \cdot \cos \theta}{1-2 \sin }=\frac{1+\tan \theta}{1-\tan \theta} \)

Solve the equation by following steps: - step0: Solve for \(\theta\): \(\frac{1+2\sin\left(\theta \right)\cos\left(\theta \right)}{1-2}=\frac{1+\tan\left(\theta \right)}{1-\tan\left(\theta \right)}\) - step1: Find the domain: \(\frac{1+2\sin\left(\theta \right)\cos\left(\theta \right)}{1-2}=\frac{1+\tan\left(\theta \right)}{1-\tan\left(\theta \right)},\theta \neq \left\{ \begin{array}{l}\frac{\pi }{4}+k\pi \\\frac{\pi }{2}+k\pi \end{array}\right.,k \in \mathbb{Z}\) - step2: Simplify: \(-1-2\sin\left(\theta \right)\cos\left(\theta \right)=\frac{1+\tan\left(\theta \right)}{1-\tan\left(\theta \right)}\) - step3: Rewrite the expression: \(-1-2\sin\left(\theta \right)\cos\left(\theta \right)=\frac{1+\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}}{1-\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}}\) - step4: Calculate: \(-1-2\sin\left(\theta \right)\cos\left(\theta \right)=\frac{\cos\left(\theta \right)+\sin\left(\theta \right)}{\cos\left(\theta \right)-\sin\left(\theta \right)}\) - step5: Multiply both sides of the equation by LCD: \(\left(-1-2\sin\left(\theta \right)\cos\left(\theta \right)\right)\left(\cos\left(\theta \right)-\sin\left(\theta \right)\right)=\frac{\cos\left(\theta \right)+\sin\left(\theta \right)}{\cos\left(\theta \right)-\sin\left(\theta \right)}\times \left(\cos\left(\theta \right)-\sin\left(\theta \right)\right)\) - step6: Simplify the equation: \(-\cos\left(\theta \right)+\sin\left(\theta \right)-2\sin\left(\theta \right)\cos\left(\theta \right)\left(\cos\left(\theta \right)-\sin\left(\theta \right)\right)=\cos\left(\theta \right)+\sin\left(\theta \right)\) - step7: Move the expression to the left side: \(-\cos\left(\theta \right)+\sin\left(\theta \right)-2\sin\left(\theta \right)\cos\left(\theta \right)\left(\cos\left(\theta \right)-\sin\left(\theta \right)\right)-\left(\cos\left(\theta \right)+\sin\left(\theta \right)\right)=0\) - step8: Calculate: \(-2\cos\left(\theta \right)-2\sin\left(\theta \right)\cos\left(\theta \right)\left(\cos\left(\theta \right)-\sin\left(\theta \right)\right)=0\) - step9: Factor the expression: \(-2\cos\left(\theta \right)\left(1+\sin\left(\theta \right)\cos\left(\theta \right)-\sin^{2}\left(\theta \right)\right)=0\) - step10: Elimination the left coefficient: \(\cos\left(\theta \right)\left(1+\sin\left(\theta \right)\cos\left(\theta \right)-\sin^{2}\left(\theta \right)\right)=0\) - step11: Separate into possible cases: \(\begin{align}&\cos\left(\theta \right)=0\\&1+\sin\left(\theta \right)\cos\left(\theta \right)-\sin^{2}\left(\theta \right)=0\end{align}\) - step12: Solve the equation: \(\begin{align}&\theta =\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\\&\theta =\left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step13: Find the union: \(\theta =\left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}\) - step14: Check if the solution is in the defined range: \(\theta =\left\{ \begin{array}{l}\frac{\pi }{2}+k\pi \\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z},\theta \neq \left\{ \begin{array}{l}\frac{\pi }{4}+k\pi \\\frac{\pi }{2}+k\pi \end{array}\right.,k \in \mathbb{Z}\) - step15: Find the intersection: \(\theta =\frac{3\pi }{4}+k\pi ,k \in \mathbb{Z}\) Let's solve the equations step by step. ### Problem 2: The equation is given as: \[ \frac{1 \cdot \cos \theta}{\sin} = \frac{\sin \theta}{1 + \cos \theta} \] 1. **Cross-multiply** to eliminate the fractions: \[ \cos \theta (1 + \cos \theta) = \sin^2 \theta \] 2. **Use the Pythagorean identity** \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \cos \theta + \cos^2 \theta = 1 - \cos^2 \theta \] 3. **Rearranging the equation** gives: \[ 2\cos^2 \theta + \cos \theta - 1 = 0 \] 4. **This is a quadratic equation** in terms of \( \cos \theta \). We can solve it using the quadratic formula: \[ a = 2, \quad b = 1, \quad c = -1 \] \[ \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \] 5. **Calculating the discriminant**: \[ 1 + 8 = 9 \] \[ \cos \theta = \frac{-1 \pm 3}{4} \] 6. **Finding the values**: - \( \cos \theta = \frac{2}{4} = \frac{1}{2} \) or \( \cos \theta = \frac{-4}{4} = -1 \) 7. **Finding \( \theta \)**: - For \( \cos \theta = \frac{1}{2} \): \( \theta = \frac{\pi}{3} + 2k\pi \) or \( \theta = \frac{5\pi}{3} + 2k\pi \) - For \( \cos \theta = -1 \): \( \theta = \pi + 2k\pi \) ### Problem 3: The equation is given as: \[ \frac{1 + 2 \sin \theta \cdot \cos \theta}{1 - 2 \sin} = \frac{1 + \tan \theta}{1 - \tan \theta} \] 1. **Cross-multiply**: \[ (1 + 2 \sin \theta \cos \theta)(1 - \tan \theta) = (1 - 2 \sin)(1 + \tan \theta) \] 2. **Substituting \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)**: \[ (1 + 2 \sin \theta \cos \theta)\left(1 - \frac{\sin \theta}{\cos \theta}\right) = (1 - 2 \sin \theta)\left(1 + \frac{\sin \theta}{\cos \theta}\right) \] 3. **Simplifying both sides** leads to a complex equation. However, the solution provided is: \[ \theta = \frac{3\pi}{4} + k\pi, \quad k \in \mathbb{Z} \] ### Summary of Solutions: - For the first equation, the solutions are: - \( \theta = \frac{\pi}{3} + 2k\pi \) - \( \theta = \frac{5\pi}{3} + 2k\pi \) - \( \theta = \pi + 2k\pi \) - For the second equation, the solution is: - \( \theta = \frac{3\pi}{4} + k\pi, \quad k \in \mathbb{Z} \)