1. Start with the given equation:
\(\csc \theta = 3 \sin \theta + \cot \theta\)
2. Use the identities \(\csc \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\):
\(\frac{1}{\sin \theta} = 3 \sin \theta + \frac{\cos \theta}{\sin \theta}\)
3. Multiply through by \(\sin \theta\) to clear the fractions:
\(1 = 3 \sin^2 \theta + \cos \theta\)
4. Use the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(1 = 3(1 - \cos^2 \theta) + \cos \theta\)
5. Simplify and rearrange to form a quadratic equation:
\(1 = 3 - 3\cos^2 \theta + \cos \theta\)
\(0 = 3\cos^2 \theta - \cos \theta - 2\)
6. Solve the quadratic equation \(3\cos^2 \theta - \cos \theta - 2 = 0\) using the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -1\), \(c = -2\):
\(\cos \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 3 \times (-2)}}{2 \times 3}\)
\(\cos \theta = \frac{1 \pm \sqrt{1 + 24}}{6}\)
\(\cos \theta = \frac{1 \pm 5}{6}\)
\(\cos \theta = 1 \text{ or } \cos \theta = -\frac{2}{3}\)
7. Since \(\cos \theta = 1\) is not in the interval \(0^\circ < \theta < 180^\circ\), we use \(\cos \theta = -\frac{2}{3}\):
\(\theta = \cos^{-1}\left(-\frac{2}{3}\right) \approx 131.8^\circ\)