First, express \(8 \sin \theta - 6 \cos \theta\) in the form \(R \sin(\theta - \alpha)\).

Using the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha),\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 8\) and \(b = -6\).

Calculate \(R\):

\(R = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-6}{8} = -\frac{3}{4}.\)

\(\alpha = \tan^{-1}\left(-\frac{3}{4}\right) \approx -36.87^\circ.\)

Thus, \(8 \sin \theta - 6 \cos \theta = 10 \sin(\theta + 36.87^\circ)\).

Now, solve the equation:

\(10 \sin(\theta + 36.87^\circ) = 7.\)

\(\sin(\theta + 36.87^\circ) = 0.7.\)

Find \(\theta + 36.87^\circ\):

\(\theta + 36.87^\circ = \sin^{-1}(0.7) \approx 44.427^\circ.\)

\(\theta = 44.427^\circ - 36.87^\circ = 7.557^\circ.\)

Considering the range \(0^\circ \leq \theta \leq 360^\circ\), find the other solution:

\(\theta = 180^\circ - 44.427^\circ - 36.87^\circ = 172.4^\circ.\)

Thus, the solutions are \(\theta = 81.3^\circ\) and \(\theta = 172.4^\circ\).