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\( \frac { \cos x } { \operatorname { cosec } + \sin x } - \frac { \cos x } { \cos x \sin x } = \operatorname { tas } 2 x \)

Ask by Hampton Black. in South Africa
Mar 15,2025

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Answer

The equation has solutions at \( x = 0 \) and \( x = \pi \).

Solution

The equation given is: \[ \frac{\cos x}{\csc x + \sin x} - \frac{\cos x}{\cos x \sin x} = \tan 2x \] Let's break this down step by step. ### Step 1: Simplify the Left Side 1. **Convert \(\csc x\)**: \[ \csc x = \frac{1}{\sin x} \] Therefore, the left side becomes: \[ \frac{\cos x}{\frac{1}{\sin x} + \sin x} - \frac{\cos x}{\cos x \sin x} \] 2. **Combine the terms in the denominator**: \[ \frac{1}{\sin x} + \sin x = \frac{1 + \sin^2 x}{\sin x} \] Thus, the first term becomes: \[ \frac{\cos x \sin x}{1 + \sin^2 x} \] 3. **Rewrite the second term**: \[ \frac{\cos x}{\cos x \sin x} = \frac{1}{\sin x} \] 4. **Combine the two fractions**: \[ \frac{\cos x \sin x}{1 + \sin^2 x} - \frac{1}{\sin x} \] ### Step 2: Find a Common Denominator The common denominator for the two fractions is \((1 + \sin^2 x) \sin x\): \[ \frac{\cos x \sin^2 x - (1 + \sin^2 x)}{(1 + \sin^2 x) \sin x} \] ### Step 3: Set the Left Side Equal to the Right Side Now we have: \[ \frac{\cos x \sin^2 x - (1 + \sin^2 x)}{(1 + \sin^2 x) \sin x} = \tan 2x \] ### Step 4: Express \(\tan 2x\) Recall that: \[ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \] ### Step 5: Solve the Equation At this point, the equation is quite complex. To solve it, we can either simplify further or use numerical methods. Given the complexity, I will attempt to solve it numerically for specific values of \(x\) or use a graphical approach. Let's try to find specific solutions for \(x\) by evaluating the left and right sides for common angles. ### Step 6: Evaluate for Specific Values Let's evaluate for \(x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\). 1. **For \(x = 0\)**: - Left: \(\frac{0}{1} - \frac{0}{0} = 0\) - Right: \(\tan 0 = 0\) 2. **For \(x = \frac{\pi}{4}\)**: - Left: \(\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}} - \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}} = \frac{\frac{\sqrt{2}}{2}}{\sqrt{2}} - 1 = 0\) - Right: \(\tan \frac{\pi}{2}\) (undefined) 3. **For \(x = \frac{\pi}{2}\)**: - Left: Undefined - Right: Undefined 4. **For \(x = \frac{3\pi}{4}\)**: - Left: \(\frac{-\frac{\sqrt{2}}{2}}{0} - \frac{-\frac{\sqrt{2}}{2}}{0} = 0\) - Right: \(\tan \frac{3\pi}{2}\) (undefined) 5. **For \(x = \pi\)**: - Left: \(0\) - Right: \(0\) ### Conclusion The equation has solutions at \(x = 0\) and \(x = \pi\). Further analysis or numerical methods may be required for other potential solutions.

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Let’s start by simplifying the given equation step by step. We have: \[ \frac{\cos x}{\csc x + \sin x} - \frac{\cos x}{\cos x \sin x} = \tan 2x \] First, recall that \(\csc x = \frac{1}{\sin x}\). So, we can rewrite the first term: \[ \frac{\cos x}{\frac{1}{\sin x} + \sin x} = \frac{\cos x}{\frac{1 + \sin^2 x}{\sin x}} = \frac{\cos x \sin x}{1 + \sin^2 x} \] Now, substituting this back into the equation gives us: \[ \frac{\cos x \sin x}{1 + \sin^2 x} - \frac{\cos x}{\cos x \sin x} = \tan 2x \] The second term simplifies to: \[ \frac{\cos x}{\cos x \sin x} = \frac{1}{\sin x} = \csc x \] Now, the equation becomes: \[ \frac{\cos x \sin x}{1 + \sin^2 x} - \csc x = \tan 2x \] Converting it back, we can express: \[ \csc x = \frac{1}{\sin x} \] Combining the fractions on the left, we can find a common denominator: \[ \frac{\cos x \sin^2 x - 1 - \sin^2 x}{(1 + \sin^2 x) \sin x} \] The final expression now can be compared to \(\tan 2x\), which is \(\frac{2 \tan x}{1 - \tan^2 x}\). By equating both sides, you can simplify further and analyze for specific x values to find solutions or necessary conditions for \(x\).

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