g) \( \frac{\tan \theta}{\tan 2 \theta}=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{1+\cos 2 \theta} \) h) \( \frac{\cos 2 \theta+\cos \theta+1}{\sin \theta+\sin 2 \theta}=\frac{1}{\tan \theta} \) i) \( \frac{\sin 2 x-\sin x}{\cos 2 x+\cos x}=\frac{\sin x}{\cos x+1} \) j) \( \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \)

Solve the equation by following steps: - step0: Solve for \(\theta\): \(\cos\left(3\theta \right)=4\cos^{3}\left(\theta \right)-3\cos\left(\theta \right)\) - step1: Rewrite the expression: \(4\cos^{3}\left(\theta \right)-3\cos\left(\theta \right)=4\cos^{3}\left(\theta \right)-3\cos\left(\theta \right)\) - step2: The statement is true: \(\theta \in \mathbb{R}\) Solve the equation \( \frac{\tan \theta}{\tan 2 \theta}=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{1+\cos 2 \theta} \). Solve the equation by following steps: - step0: Solve for \(\theta\): \(\frac{\tan\left(\theta \right)}{\tan\left(2\theta \right)}=\frac{\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)}{1+\cos\left(2\theta \right)}\) - step1: Find the domain: \(\frac{\tan\left(\theta \right)}{\tan\left(2\theta \right)}=\frac{\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)}{1+\cos\left(2\theta \right)},\theta \neq \frac{k\pi }{4},k \in \mathbb{Z}\) - step2: Calculate: \(\frac{\tan\left(\theta \right)}{\tan\left(2\theta \right)}=\frac{\cos\left(2\theta \right)}{1+\cos\left(2\theta \right)}\) - step3: Rewrite the expression: \(\tan^{-1}\left(2\theta \right)\tan\left(\theta \right)\) - step4: Express with a positive exponent: \(\frac{1}{\tan\left(2\theta \right)}\times \tan\left(\theta \right)\) - step5: Rewrite the expression: \(\frac{\tan\left(\theta \right)}{\tan\left(2\theta \right)}\) - step6: Rewrite the expression: \(\frac{\tan\left(\theta \right)}{\tan\left(2\theta \right)}=\left(1+\cos\left(2\theta \right)\right)^{-1}\cos\left(2\theta \right)\) - step7: Rewrite the expression: \(\frac{\tan\left(\theta \right)}{\frac{2\tan\left(\theta \right)}{1-\tan^{2}\left(\theta \right)}}=\left(1+2\cos^{2}\left(\theta \right)-1\right)^{-1}\left(2\cos^{2}\left(\theta \right)-1\right)\) - step8: Reduce the fraction: \(\frac{1-\tan^{2}\left(\theta \right)}{2}=\left(1+2\cos^{2}\left(\theta \right)-1\right)^{-1}\left(2\cos^{2}\left(\theta \right)-1\right)\) - step9: Expand the expression: \(\frac{1-\tan^{2}\left(\theta \right)}{2}=2\left(1+2\cos^{2}\left(\theta \right)-1\right)^{-1}\cos^{2}\left(\theta \right)-\left(1+2\cos^{2}\left(\theta \right)-1\right)^{-1}\) - step10: Rewrite the expression: \(\frac{1-\left(\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}\right)^{2}}{2}=2\times \frac{1}{1+2\cos^{2}\left(\theta \right)-1}\times \cos^{2}\left(\theta \right)-\frac{1}{1+2\cos^{2}\left(\theta \right)-1}\) - step11: Simplify: \(\frac{1-\left(\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}\right)^{2}}{2}=2\times \frac{1}{1+2\cos^{2}\left(\theta \right)-1}\times \cos^{2}\left(\theta \right)-\frac{1}{2\cos^{2}\left(\theta \right)}\) - step12: Expand the expression: \(\frac{1-\frac{\sin^{2}\left(\theta \right)}{\cos^{2}\left(\theta \right)}}{2}=2\times \frac{1}{1+2\cos^{2}\left(\theta \right)-1}\times \cos^{2}\left(\theta \right)-\frac{1}{2\cos^{2}\left(\theta \right)}\) - step13: Simplify: \(\frac{1-\frac{\sin^{2}\left(\theta \right)}{\cos^{2}\left(\theta \right)}}{2}=2\left(1+2\cos^{2}\left(\theta \right)-1\right)^{-1}\cos^{2}\left(\theta \right)-\frac{1}{2\cos^{2}\left(\theta \right)}\) - step14: Calculate: \(\frac{\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)}{2\cos^{2}\left(\theta \right)}=2\left(1+2\cos^{2}\left(\theta \right)-1\right)^{-1}\cos^{2}\left(\theta \right)-\frac{1}{2\cos^{2}\left(\theta \right)}\) - step15: Rewrite the expression: \(\frac{\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)}{2\cos^{2}\left(\theta \right)}=\frac{2\cos^{2}\left(\theta \right)}{1+2\cos^{2}\left(\theta \right)-1}-\frac{1}{2\cos^{2}\left(\theta \right)}\) - step16: Multiply both sides of the equation by LCD: \(\frac{\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)}{2\cos^{2}\left(\theta \right)}\times 2\cos^{2}\left(\theta \right)=\left(\frac{2\cos^{2}\left(\theta \right)}{1+2\cos^{2}\left(\theta \right)-1}-\frac{1}{2\cos^{2}\left(\theta \right)}\right)\times 2\cos^{2}\left(\theta \right)\) - step17: Simplify the equation: \(\cos^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)=2\cos^{2}\left(\theta \right)-1\) - step18: Rewrite the expression: \(2\cos^{2}\left(\theta \right)-1=2\cos^{2}\left(\theta \right)-1\) - step19: Cancel equal terms: \(0=0\) - step20: The statement is true: \(\theta \in \mathbb{R}\) - step21: Check if the solution is in the defined range: \(\theta \in \mathbb{R},\theta \neq \frac{k\pi }{4},k \in \mathbb{Z}\) - step22: Find the intersection: \(\theta \neq \frac{k\pi }{4},k \in \mathbb{Z}\) Solve the equation \( \frac{\cos 2 \theta+\cos \theta+1}{\sin \theta+\sin 2 \theta}=\frac{1}{\tan \theta} \). Solve the equation by following steps: - step0: Solve for \(\theta\): \(\frac{\cos\left(2\theta \right)+\cos\left(\theta \right)+1}{\sin\left(\theta \right)+\sin\left(2\theta \right)}=\frac{1}{\tan\left(\theta \right)}\) - step1: Find the domain: \(\frac{\cos\left(2\theta \right)+\cos\left(\theta \right)+1}{\sin\left(\theta \right)+\sin\left(2\theta \right)}=\frac{1}{\tan\left(\theta \right)},\theta \neq \left\{ \begin{array}{l}\frac{2k\pi }{3}\\\frac{k\pi }{2}\end{array}\right.,k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\left(\cos\left(2\theta \right)+\cos\left(\theta \right)+1\right)\left(\sin\left(\theta \right)+\sin\left(2\theta \right)\right)^{-1}\) - step3: Express with a positive exponent: \(\left(\cos\left(2\theta \right)+\cos\left(\theta \right)+1\right)\times \frac{1}{\sin\left(\theta \right)+\sin\left(2\theta \right)}\) - step4: Rewrite the expression: \(\frac{\cos\left(2\theta \right)+\cos\left(\theta \right)+1}{\sin\left(\theta \right)+\sin\left(2\theta \right)}\) - step5: Rewrite the expression: \(\frac{\cos\left(2\theta \right)+\cos\left(\theta \right)+1}{\sin\left(\theta \right)+\sin\left(2\theta \right)}=\tan^{-1}\left(\theta \right)\) - step6: Rewrite the expression: \(\frac{2\cos^{2}\left(\theta \right)-1+\cos\left(\theta \right)+1}{\sin\left(\theta \right)+2\sin\left(\theta \right)\cos\left(\theta \right)}=\tan^{-1}\left(\theta \right)\) - step7: Reduce the fraction: \(\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}=\tan^{-1}\left(\theta \right)\) - step8: Rewrite the expression: \(\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}=\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}\) - step9: The statement is true: \(\theta \in \mathbb{R}\) - step10: Check if the solution is in the defined range: \(\theta \in \mathbb{R},\theta \neq \left\{ \begin{array}{l}\frac{2k\pi }{3}\\\frac{k\pi }{2}\end{array}\right.,k \in \mathbb{Z}\) - step11: Find the intersection: \(\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\\frac{2k\pi }{3}\end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( \frac{\sin 2 x-\sin x}{\cos 2 x+\cos x}=\frac{\sin x}{\cos x+1} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\sin\left(2x\right)-\sin\left(x\right)}{\cos\left(2x\right)+\cos\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)+1}\) - step1: Find the domain: \(\frac{\sin\left(2x\right)-\sin\left(x\right)}{\cos\left(2x\right)+\cos\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)+1},x\neq \frac{\pi }{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\left(\sin\left(2x\right)-\sin\left(x\right)\right)\left(\cos\left(2x\right)+\cos\left(x\right)\right)^{-1}\) - step3: Express with a positive exponent: \(\left(\sin\left(2x\right)-\sin\left(x\right)\right)\times \frac{1}{\cos\left(2x\right)+\cos\left(x\right)}\) - step4: Rewrite the expression: \(\frac{\sin\left(2x\right)-\sin\left(x\right)}{\cos\left(2x\right)+\cos\left(x\right)}\) - step5: Rewrite the expression: \(\frac{\sin\left(2x\right)-\sin\left(x\right)}{\cos\left(2x\right)+\cos\left(x\right)}=\left(\cos\left(x\right)+1\right)^{-1}\sin\left(x\right)\) - step6: Rewrite the expression: \(\frac{2\sin\left(x\right)\cos\left(x\right)-\sin\left(x\right)}{2\cos^{2}\left(x\right)-1+\cos\left(x\right)}=\left(\cos\left(x\right)+1\right)^{-1}\sin\left(x\right)\) - step7: Reduce the fraction: \(\frac{\sin\left(x\right)}{\cos\left(x\right)+1}=\left(\cos\left(x\right)+1\right)^{-1}\sin\left(x\right)\) - step8: Rewrite the expression: \(\frac{\sin\left(x\right)}{\cos\left(x\right)+1}=\frac{\sin\left(x\right)}{\cos\left(x\right)+1}\) - step9: Cross multiply: \(\sin\left(x\right)\left(\cos\left(x\right)+1\right)=\left(\cos\left(x\right)+1\right)\sin\left(x\right)\) - step10: Move the expression to the left side: \(\sin\left(x\right)\left(\cos\left(x\right)+1\right)-\left(\cos\left(x\right)+1\right)\sin\left(x\right)=0\) - step11: Calculate: \(0=0\) - step12: The statement is true: \(x \in \mathbb{R}\) - step13: Check if the solution is in the defined range: \(x \in \mathbb{R},x\neq \frac{\pi }{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\) - step14: Find the intersection: \(x\neq \frac{\pi }{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\) Let's analyze and solve each of the equations step by step. ### g) The equation is: \[ \frac{\tan \theta}{\tan 2 \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{1 + \cos 2\theta} \] **Solution:** The solution indicates that: \[ \theta \neq \frac{k\pi}{4}, k \in \mathbb{Z} \] This means that \(\theta\) cannot take values that are multiples of \(\frac{\pi}{4}\). ### h) The equation is: \[ \frac{\cos 2\theta + \cos \theta + 1}{\sin \theta + \sin 2\theta} = \frac{1}{\tan \theta} \] **Solution:** The solution indicates that: \[ \theta \neq \left\{ \frac{k\pi}{2}, \frac{2k\pi}{3} \right., k \in \mathbb{Z} \] This means that \(\theta\) cannot take values that are multiples of \(\frac{\pi}{2}\) or \(\frac{2\pi}{3}\). ### i) The equation is: \[ \frac{\sin 2x - \sin x}{\cos 2x + \cos x} = \frac{\sin x}{\cos x + 1} \] **Solution:** The solution indicates that: \[ x \neq \frac{\pi}{3} + \frac{2k\pi}{3}, k \in \mathbb{Z} \] This means that \(x\) cannot take values that are of the form \(\frac{\pi}{3} + \frac{2k\pi}{3}\). ### j) The equation is: \[ \cos 3\theta = 4\cos^3 \theta - 3\cos \theta \] **Solution:** The solution indicates that: \[ \theta \in \mathbb{R} \] This means that \(\theta\) can take any real value. ### Summary of Results: - **g)** \(\theta \neq \frac{k\pi}{4}, k \in \mathbb{Z}\) - **h)** \(\theta \neq \left\{ \frac{k\pi}{2}, \frac{2k\pi}{3} \right., k \in \mathbb{Z}\) - **i)** \(x \neq \frac{\pi}{3} + \frac{2k\pi}{3}, k \in \mathbb{Z}\) - **j)** \(\theta \in \mathbb{R}\) (all real solutions)