(3) \( \frac{\cos ^{2} x}{1+\sin x}=1-\tan x \cos x \)

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 \( k \in \mathbb{Z} \).