To express \(\sec \theta = 3 \cos \theta + \tan \theta\) as a quadratic in \(\sin \theta\), we start by using trigonometric identities:

\(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Substitute these into the equation:

\(\frac{1}{\cos \theta} = 3 \cos \theta + \frac{\sin \theta}{\cos \theta}\)

Multiply through by \(\cos \theta\) to clear the fractions:

\(1 = 3 \cos^2 \theta + \sin \theta\)

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(1 = 3(1 - \sin^2 \theta) + \sin \theta\)

Expand and rearrange:

\(1 = 3 - 3 \sin^2 \theta + \sin \theta\)

\(3 \sin^2 \theta - \sin \theta - 2 = 0\)

This is a quadratic equation in \(\sin \theta\). Let \(x = \sin \theta\), then:

\(3x^2 - x - 2 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = -1\), \(c = -2\):

\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3}\)

\(x = \frac{1 \pm \sqrt{1 + 24}}{6}\)

\(x = \frac{1 \pm 5}{6}\)

\(x = 1 \text{ or } x = -\frac{2}{3}\)

Since \(x = \sin \theta\), and \(-1 \leq \sin \theta \leq 1\), only \(x = -\frac{2}{3}\) is valid in the interval \(-90^\circ < \theta < 90^\circ\).

Thus, \(\sin \theta = -\frac{2}{3}\).

Find \(\theta\):

\(\theta = \arcsin\left(-\frac{2}{3}\right) \approx -41.8^\circ\)