Find the general solution 1. \( 1-\sin 2 x=9 \cos ^{2} x \) \( 22 \cos 2 x=-1 \) \( 3 \quad 5 \sin x+1=4 / \sin x \) 4 \( \sin \theta=\cos \theta \) \( 5 \sin ^{2} \theta=1 \)

Check the equality by following steps: - step0: Check the equality is true or false: \(5\frac{1}{2}\times \left(1-\frac{1}{2}\right)=1\) - step1: Simplify: \(\frac{11}{4}=1\) - step2: Multiply both sides of the equation by LCD: \(\frac{11}{4}\times 4=1\times 4\) - step3: Simplify the equation: \(11=4\) - step4: Check the equality: \(\textrm{false}\) Solve the equation \( 22 \frac{1 - \frac{1}{2}(1 - \frac{1}{2})}{1 + \frac{1}{2}(1 + \frac{1}{2})} = -1 \). Check the equality by following steps: - step0: Check the equality is true or false: \(22\times \frac{1-\frac{1}{2}\left(1-\frac{1}{2}\right)}{1+\frac{1}{2}\left(1+\frac{1}{2}\right)}=-1\) - step1: Simplify: \(\frac{66}{7}=-1\) - step2: Multiply both sides of the equation by LCD: \(\frac{66}{7}\times 7=-7\) - step3: Simplify the equation: \(66=-7\) - step4: Check the equality: \(\textrm{false}\) Solve the equation \( 5 \frac{5 \frac{1}{2}(1 - \frac{1}{2})}{1 + \frac{1}{2}(1 + \frac{1}{2})} + 1 = 4 / \frac{1}{2}(1 - \frac{1}{2}) \). Check the equality by following steps: - step0: Check the equality is true or false: \(\left(5\times \frac{5\frac{1}{2}\times \left(1-\frac{1}{2}\right)}{1+\frac{1}{2}\left(1+\frac{1}{2}\right)}\right)+1=\frac{4}{\frac{1}{2}\left(1-\frac{1}{2}\right)}\) - step1: Simplify: \(\frac{62}{7}=16\) - step2: Multiply both sides of the equation by LCD: \(\frac{62}{7}\times 7=16\times 7\) - step3: Simplify the equation: \(62=112\) - step4: Check the equality: \(\textrm{false}\) Solve the equation \( \frac{1}{2}(1 - \frac{1}{2}) = \frac{1}{2}(1 - \frac{1}{2}) \). Check the equality by following steps: - step0: Check the equality is true or false: \(\frac{1}{2}\left(1-\frac{1}{2}\right)=\frac{1}{2}\left(1-\frac{1}{2}\right)\) - step1: Simplify: \(\frac{1}{4}=\frac{1}{4}\) - step2: Check the equality: \(\textrm{true}\) Solve the equation \( 1 - \frac{1 - \frac{1}{2}(1 - \frac{1}{2})}{1 + \frac{1}{2}(1 + \frac{1}{2})} = 9 \frac{1 + \frac{1}{2}(1 - \frac{1}{2})}{1 + \frac{1}{2}(1 + \frac{1}{2})} \). Check the equality by following steps: - step0: Check the equality is true or false: \(1-\frac{1-\frac{1}{2}\left(1-\frac{1}{2}\right)}{1+\frac{1}{2}\left(1+\frac{1}{2}\right)}=9\times \frac{1+\frac{1}{2}\left(1-\frac{1}{2}\right)}{1+\frac{1}{2}\left(1+\frac{1}{2}\right)}\) - step1: Simplify: \(\frac{4}{7}=\frac{45}{7}\) - step2: Calculate: \(0.\dot{5}7142\dot{8}=\frac{45}{7}\) - step3: Calculate: \(0.\dot{5}7142\dot{8}=6.\dot{4}2857\dot{1}\) - step4: Check the equality: \(\textrm{false}\) Let's solve each equation step by step. 1. **Equation:** \( 1 - \sin 2x = 9 \cos^2 x \) We can use the identity \( \sin 2x = 2 \sin x \cos x \) and \( \cos^2 x = 1 - \sin^2 x \) to rewrite the equation. However, it seems that the solution returned "false," indicating that there may not be a valid solution. 2. **Equation:** \( 22 \cos 2x = -1 \) Using the identity \( \cos 2x = 2 \cos^2 x - 1 \), we can rewrite this as: \[ 22(2 \cos^2 x - 1) = -1 \] Simplifying gives: \[ 44 \cos^2 x - 22 = -1 \implies 44 \cos^2 x = 21 \implies \cos^2 x = \frac{21}{44} \] Thus, \( \cos x = \pm \sqrt{\frac{21}{44}} \). 3. **Equation:** \( 5 \sin x + 1 = \frac{4}{\sin x} \) Multiplying through by \( \sin x \) gives: \[ 5 \sin^2 x + \sin x - 4 = 0 \] This is a quadratic equation in \( \sin x \). We can solve it using the quadratic formula: \[ \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 5 \cdot (-4)}}{2 \cdot 5} \] Simplifying gives: \[ \sin x = \frac{-1 \pm \sqrt{81}}{10} = \frac{-1 \pm 9}{10} \] Thus, \( \sin x = \frac{8}{10} = 0.8 \) or \( \sin x = \frac{-10}{10} = -1 \). 4. **Equation:** \( \sin \theta = \cos \theta \) This implies \( \tan \theta = 1 \), so: \[ \theta = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \] 5. **Equation:** \( 5 \sin^2 \theta = 1 \) This simplifies to: \[ \sin^2 \theta = \frac{1}{5} \implies \sin \theta = \pm \frac{1}{\sqrt{5}} \] Thus, the solutions are: \[ \theta = \arcsin\left(\frac{1}{\sqrt{5}}\right) + 2n\pi \quad \text{and} \quad \theta = \arcsin\left(-\frac{1}{\sqrt{5}}\right) + 2n\pi, \quad n \in \mathbb{Z} \] ### Summary of Solutions: 1. No valid solution for \( 1 - \sin 2x = 9 \cos^2 x \). 2. \( \cos x = \pm \sqrt{\frac{21}{44}} \). 3. \( \sin x = 0.8 \) or \( \sin x = -1 \). 4. \( \theta = \frac{\pi}{4} + n\pi \). 5. \( \theta = \arcsin\left(\frac{1}{\sqrt{5}}\right) + 2n\pi \) and \( \theta = \arcsin\left(-\frac{1}{\sqrt{5}}\right) + 2n\pi \).