3) \( 5.52+3 \csc \theta=5+\frac{11}{4} \cdot \csc \theta \)

Solve the equation \( 5.52+3\csc(\theta)=5+\frac{11}{4}\csc(\theta) \). Solve the equation by following steps: - step0: Solve for \(\theta\): \(5.52+3\csc\left(\theta \right)=5+\frac{11}{4}\csc\left(\theta \right)\) - step1: Find the domain: \(5.52+3\csc\left(\theta \right)=5+\frac{11}{4}\csc\left(\theta \right),\theta \neq k\pi ,k \in \mathbb{Z}\) - step2: Move the expression to the left side: \(5.52+3\csc\left(\theta \right)-\left(5+\frac{11}{4}\csc\left(\theta \right)\right)=0\) - step3: Calculate: \(0.52+\frac{1}{4}\csc\left(\theta \right)=0\) - step4: Move the constant to the right side: \(\frac{1}{4}\csc\left(\theta \right)=0-0.52\) - step5: Remove 0: \(\frac{1}{4}\csc\left(\theta \right)=-0.52\) - step6: Multiply by the reciprocal: \(\frac{1}{4}\csc\left(\theta \right)\times 4=-0.52\times 4\) - step7: Multiply: \(\csc\left(\theta \right)=-2.08\) - step8: Use the inverse trigonometric function: \(\theta =\operatorname{arccsc}\left(-2.08\right)\) - step9: Calculate: \(\begin{align}&\theta =-\operatorname{arccsc}\left(2.08\right)\\&\theta =\operatorname{arccsc}\left(2.08\right)+\pi \end{align}\) - step10: Add the period: \(\begin{align}&\theta =-\operatorname{arccsc}\left(2.08\right)+2k\pi ,k \in \mathbb{Z}\\&\theta =\operatorname{arccsc}\left(2.08\right)+\pi +2k\pi ,k \in \mathbb{Z}\end{align}\) - step11: Find the union: \(\theta =\left\{ \begin{array}{l}-\operatorname{arccsc}\left(2.08\right)+2k\pi \\\operatorname{arccsc}\left(2.08\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\) - step12: Check if the solution is in the defined range: \(\theta =\left\{ \begin{array}{l}-\operatorname{arccsc}\left(2.08\right)+2k\pi \\\operatorname{arccsc}\left(2.08\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z},\theta \neq k\pi ,k \in \mathbb{Z}\) - step13: Find the intersection: \(\theta =\left\{ \begin{array}{l}-\operatorname{arccsc}\left(2.08\right)+2k\pi \\\operatorname{arccsc}\left(2.08\right)+\pi +2k\pi \end{array}\right.,k \in \mathbb{Z}\) - step14: Simplify: \(\theta \approx \left\{ \begin{array}{l}-\operatorname{arccsc}\left(2.08\right)+2k\pi \\3.643124+2k\pi \end{array}\right.,k \in \mathbb{Z}\) The solutions to the equation \(5.52+3\csc(\theta)=5+\frac{11}{4}\csc(\theta)\) are: \[ \theta \approx \left\{ \begin{array}{l} -\operatorname{arccsc}(2.08)+2k\pi \\ 3.643124+2k\pi \end{array} \right., k \in \mathbb{Z} \] \[ \theta \approx \left\{ \begin{array}{l} -0.501532+2k\pi \\ 3.643124+2k\pi \end{array} \right., k \in \mathbb{Z} \]