Pregunta

\( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \)

Ask by Martinez Bush. in South Africa
Mar 11,2025

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The solution to the equation is all angles \( \alpha \) except \( \alpha = \frac{3\pi}{2} + 2k\pi \) for any integer \( k \).

Solución

Solve the equation \( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \). Solve the equation by following steps: - step0: Solve for \(\alpha\): \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\frac{2\sin\left(\alpha \right)-1}{\sin\left(\alpha \right)-2}\) - step1: Find the domain: \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\frac{2\sin\left(\alpha \right)-1}{\sin\left(\alpha \right)-2},\alpha \neq \frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\left(\cos\left(2\alpha \right)-\sin\left(\alpha \right)\right)\left(1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)\right)^{-1}\) - step3: Express with a positive exponent: \(\left(\cos\left(2\alpha \right)-\sin\left(\alpha \right)\right)\times \frac{1}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}\) - step4: Rewrite the expression: \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}\) - step5: Rewrite the expression: \(\frac{\cos\left(2\alpha \right)-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\left(2\sin\left(\alpha \right)-1\right)\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step6: Rewrite the expression: \(\frac{2\cos^{2}\left(\alpha \right)-1-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=\left(2\sin\left(\alpha \right)-1\right)\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step7: Expand the expression: \(\frac{2\cos^{2}\left(\alpha \right)-1-\sin\left(\alpha \right)}{1+\sin\left(\alpha \right)+\cos^{2}\left(\alpha \right)}=2\sin\left(\alpha \right)\left(\sin\left(\alpha \right)-2\right)^{-1}-\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step8: Rewrite the expression: \(\frac{1-2\sin\left(\alpha \right)}{2-\sin\left(\alpha \right)}=2\sin\left(\alpha \right)\left(\sin\left(\alpha \right)-2\right)^{-1}-\left(\sin\left(\alpha \right)-2\right)^{-1}\) - step9: Rewrite the expression: \(\frac{1-2\sin\left(\alpha \right)}{2-\sin\left(\alpha \right)}=\frac{2\sin\left(\alpha \right)}{\sin\left(\alpha \right)-2}-\frac{1}{\sin\left(\alpha \right)-2}\) - step10: Multiply both sides of the equation by LCD: \(\frac{1-2\sin\left(\alpha \right)}{2-\sin\left(\alpha \right)}\times \left(-\sin\left(\alpha \right)+2\right)=\left(\frac{2\sin\left(\alpha \right)}{\sin\left(\alpha \right)-2}-\frac{1}{\sin\left(\alpha \right)-2}\right)\left(-\sin\left(\alpha \right)+2\right)\) - step11: Simplify the equation: \(1-2\sin\left(\alpha \right)=-2\sin\left(\alpha \right)+1\) - step12: Move the expression to the left side: \(1-2\sin\left(\alpha \right)-\left(-2\sin\left(\alpha \right)+1\right)=0\) - step13: Calculate: \(0=0\) - step14: The statement is true: \(\alpha \in \mathbb{R}\) - step15: Check if the solution is in the defined range: \(\alpha \in \mathbb{R},\alpha \neq \frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\) - step16: Find the intersection: \(\alpha \neq \frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\) The solution to the equation \( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \) is \( \alpha \neq \frac{3\pi}{2}+2k\pi \) where \( k \in \mathbb{Z} \).

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Mind Expander

To solve the equation \( \frac{\cos 2 \alpha-\sin \alpha}{1+\sin \alpha+\cos ^{2} \alpha}=\frac{2 \sin \alpha-1}{\sin \alpha-2} \), we can begin with some trigonometric identities and transformations. First, recall that \( \cos 2\alpha = 1 - 2\sin^2\alpha \). So we can substitute this into the equation: \[ \frac{(1 - 2\sin^2 \alpha) - \sin \alpha}{1+\sin \alpha + \cos^2 \alpha} \] Now, observe that \( \cos^2 \alpha = 1 - \sin^2 \alpha \), hence \( 1 + \sin \alpha + \cos^2 \alpha = 1 + \sin \alpha + (1 - \sin^2 \alpha) = 2 + \sin \alpha - \sin^2 \alpha \). Thus, the left side becomes: \[ \frac{1 - 2\sin^2 \alpha - \sin \alpha}{2 + \sin \alpha - \sin^2 \alpha} \] We can now work on simplifying both sides. For the right side, notice that if we rearrange, we will have: \[ \frac{2\sin\alpha - 1}{\sin\alpha - 2} \] Next, we can cross-multiply to eliminate the fractions: \[ (1 - 2\sin^2 \alpha - \sin \alpha)(\sin \alpha - 2) = (2\sin \alpha - 1)(2 + \sin \alpha - \sin^2 \alpha) \] Expanding both sides will yield a polynomial equation where you can collect all terms to one side, simplify, and then solve for \(\alpha\). However, a quicker numerical or graphical approach can also be used if feasible, by plotting both sides of the original equation to find potential solutions for \( \alpha \). Have fun solving it, and remember to check for extraneous solutions that might arise when squaring or simplifying!

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