Answer
The solution to the equation \( \frac{\cos ^{2}x}{1+\sin x}=1-\tan x\cos x \) is all real numbers except \( x = \frac{\pi}{2} + k\pi \) where \( k \) is any integer.
Solution
Solve the equation \( \frac{\cos ^{2}x}{1+\sin x}=1-\tan x\cos x \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\frac{\cos^{2}\left(x\right)}{1+\sin\left(x\right)}=1-\tan\left(x\right)\cos\left(x\right)\)
- step1: Find the domain:
\(\frac{\cos^{2}\left(x\right)}{1+\sin\left(x\right)}=1-\tan\left(x\right)\cos\left(x\right),x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\)
- step2: Rewrite the expression:
\(\frac{\cos^{2}\left(x\right)}{1+\sin\left(x\right)}=1-\frac{\sin\left(x\right)}{\cos\left(x\right)}\times \cos\left(x\right)\)
- step3: Rewrite the expression:
\(\frac{\cos^{2}\left(x\right)}{1+\sin\left(x\right)}=1-\frac{\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)}\)
- step4: Multiply both sides of the equation by LCD:
\(\frac{\cos^{2}\left(x\right)}{1+\sin\left(x\right)}\times \left(1+\sin\left(x\right)\right)\cos\left(x\right)=\left(1-\frac{\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)}\right)\left(1+\sin\left(x\right)\right)\cos\left(x\right)\)
- step5: Simplify the equation:
\(\cos^{3}\left(x\right)=\left(1+\sin\left(x\right)\right)\cos\left(x\right)-\sin\left(x\right)\cos\left(x\right)\left(1+\sin\left(x\right)\right)\)
- step6: Move the expression to the left side:
\(\cos^{3}\left(x\right)-\left(\left(1+\sin\left(x\right)\right)\cos\left(x\right)-\sin\left(x\right)\cos\left(x\right)\left(1+\sin\left(x\right)\right)\right)=0\)
- step7: Calculate:
\(\cos^{3}\left(x\right)+\left(-1-\sin\left(x\right)\right)\cos\left(x\right)+\sin\left(x\right)\cos\left(x\right)\left(1+\sin\left(x\right)\right)=0\)
- step8: Factor the expression:
\(\cos\left(x\right)\left(\cos^{2}\left(x\right)-1+\sin^{2}\left(x\right)\right)=0\)
- step9: Separate into possible cases:
\(\begin{align}&\cos\left(x\right)=0\\&\cos^{2}\left(x\right)-1+\sin^{2}\left(x\right)=0\end{align}\)
- step10: Solve the equation:
\(\begin{align}&x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\\&x \in \mathbb{R}\end{align}\)
- step11: Find the union:
\(x \in \mathbb{R}\)
- step12: Check if the solution is in the defined range:
\(x \in \mathbb{R},x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\)
- step13: Find the intersection:
\(x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\)
The solution to the equation \( \frac{\cos ^{2}x}{1+\sin x}=1-\tan x\cos x \) is \( x \neq \frac{\pi}{2}+k\pi \) where \( k \in \mathbb{Z} \).
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