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5) \( \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=\tan 2 x \)

Ask by Morrison Spencer. in South Africa
Mar 15,2025

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Answer

The solution to the equation is all real numbers except \( x = \frac{\pi}{4} + \frac{k\pi}{2} \) for any integer \( k \).

Solution

Solve the equation \( \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=\tan 2x \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{\cos\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}=\tan\left(2x\right)\) - step1: Find the domain: \(\frac{\cos\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}=\tan\left(2x\right),x\neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\frac{\cos\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}=\frac{2\tan\left(x\right)}{1-\tan^{2}\left(x\right)}\) - step3: Rewrite the expression: \(\frac{\cos\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}=\frac{2\times \frac{\sin\left(x\right)}{\cos\left(x\right)}}{1-\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)^{2}}\) - step4: Expand the expression: \(\frac{\cos\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}=\frac{\frac{2\sin\left(x\right)}{\cos\left(x\right)}}{1-\frac{\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}}\) - step5: Calculate: \(\frac{\cos\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}=\frac{2\sin\left(x\right)\cos\left(x\right)}{\cos^{2}\left(x\right)-\sin^{2}\left(x\right)}\) - step6: Multiply both sides of the equation by LCD: \(\left(\frac{\cos\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}\right)\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)=\frac{2\sin\left(x\right)\cos\left(x\right)}{\cos^{2}\left(x\right)-\sin^{2}\left(x\right)}\times \left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)\) - step7: Simplify the equation: \(2\sin\left(x\right)\cos\left(x\right)=2\sin\left(x\right)\cos\left(x\right)\) - step8: Move the expression to the left side: \(2\sin\left(x\right)\cos\left(x\right)-2\sin\left(x\right)\cos\left(x\right)=0\) - step9: Calculate: \(0=0\) - step10: The statement is true: \(x \in \mathbb{R}\) - step11: Check if the solution is in the defined range: \(x \in \mathbb{R},x\neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\) - step12: Find the intersection: \(x\neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\) The solution to the equation \( \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=\tan 2x \) is \( x \neq \frac{\pi}{4} + \frac{k\pi}{2}, k \in \mathbb{Z} \).

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To solve the equation \( \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=\tan 2 x \), we can start by simplifying the left-hand side. Combine the fractions: \[ \frac{\cos x}{\cos x - \sin x} - \frac{\cos x}{\cos x + \sin x} = \frac{\cos x(\cos x + \sin x) - \cos x(\cos x - \sin x)}{(\cos x - \sin x)(\cos x + \sin x)} \] Now, simplify the numerator: \[ \cos x(\cos x + \sin x) - \cos x(\cos x - \sin x) = \cos x \sin x + \cos x \sin x = 2 \cos x \sin x \] The denominator is: \[ (\cos x - \sin x)(\cos x + \sin x) = \cos^2 x - \sin^2 x = \cos 2x \] So, we can rewrite the left-hand side: \[ \frac{2 \cos x \sin x}{\cos 2x} \] Noticing that \( 2 \cos x \sin x = \sin 2x \), we have: \[ \frac{\sin 2x}{\cos 2x} = \tan 2x \] Thus, the left side simplifies to \( \tan 2x \), which matches the right side of the equation. Therefore, the equation holds true and is satisfied for all \( x \) for which \( \cos x \neq \sin x \) and \( \cos 2x \neq 0 \) (to avoid division by zero).

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