38. \( \sec x+\tan x=1 \)

Solve the equation \( \sec(x)+\tan(x)=1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sec\left(x\right)+\tan\left(x\right)=1\) - step1: Find the domain: \(\sec\left(x\right)+\tan\left(x\right)=1,x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step2: Rewrite the expression: \(\frac{1}{\cos\left(x\right)}+\frac{\sin\left(x\right)}{\cos\left(x\right)}=1\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{1}{\cos\left(x\right)}+\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)\cos\left(x\right)=1\times \cos\left(x\right)\) - step4: Simplify the equation: \(1+\sin\left(x\right)=\cos\left(x\right)\) - step5: Raise both sides to the \(2\)-th power\(:\) \(\left(1+\sin\left(x\right)\right)^{2}=\cos^{2}\left(x\right)\) - step6: Calculate: \(1+2\sin\left(x\right)+\sin^{2}\left(x\right)=\cos^{2}\left(x\right)\) - step7: Rewrite the expression: \(1+2\sin\left(x\right)+\sin^{2}\left(x\right)=1-\sin^{2}\left(x\right)\) - step8: Move the expression to the left side: \(1+2\sin\left(x\right)+\sin^{2}\left(x\right)-\left(1-\sin^{2}\left(x\right)\right)=0\) - step9: Calculate: \(2\sin\left(x\right)+2\sin^{2}\left(x\right)=0\) - step10: Factor the expression: \(2\sin\left(x\right)\left(1+\sin\left(x\right)\right)=0\) - step11: Calculate: \(\sin\left(x\right)\left(1+\sin\left(x\right)\right)=0\) - step12: Separate into possible cases: \(\begin{align}&\sin\left(x\right)=0\\&1+\sin\left(x\right)=0\end{align}\) - step13: Solve the equation: \(\begin{align}&x=\left\{ \begin{array}{l}2k\pi \\\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step14: Find the union: \(x=\left\{ \begin{array}{l}2k\pi \\\pi +2k\pi \\\frac{3\pi }{2}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) - step15: Check the solution: \(x=2k\pi ,k \in \mathbb{Z}\) - step16: Check if the solution is in the defined range: \(x=2k\pi ,k \in \mathbb{Z},x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step17: Find the intersection: \(x=2k\pi ,k \in \mathbb{Z}\) The solution to the equation \( \sec x + \tan x = 1 \) is \( x = 2k\pi \), where \( k \) is an integer.