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