1. **State the problem:** Solve the trigonometric equation $3 \sin \theta - 4 \cos \theta = 2$ for $\theta$. 2. **Formula and approach:** We use the identity that any expression of the form $a \sin \theta + b \cos \theta$ can be rewritten as $R \sin(\theta + \alpha)$ where $R = \sqrt{a^2 + b^2}$ and $\alpha$ satisfies $\cos \alpha = \frac{a}{R}$, $\sin \alpha = \frac{b}{R}$. 3. **Calculate $R$:** $$ R = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$ 4. **Find $\alpha$:** $$ \cos \alpha = \frac{3}{5}, \quad \sin \alpha = \frac{-4}{5} $$ Since $\sin \alpha$ is negative and $\cos \alpha$ positive, $\alpha$ is in the fourth quadrant. 5. **Rewrite the equation:** $$ 3 \sin \theta - 4 \cos \theta = 5 \sin(\theta - \alpha) = 2 $$ 6. **Solve for $\sin(\theta - \alpha)$:** $$ \sin(\theta - \alpha) = \frac{2}{5} $$ 7. **Find general solutions:** $$ \theta - \alpha = \sin^{-1}\left(\frac{2}{5}\right) \quad \text{or} \quad \theta - \alpha = \pi - \sin^{-1}\left(\frac{2}{5}\right) $$ 8. **Calculate $\sin^{-1}(2/5)$ approximately:** $$ \sin^{-1}\left(\frac{2}{5}\right) \approx 0.4115 $$ 9. **Recall $\alpha$:** $$ \alpha = \arctan\left(\frac{-4}{3}\right) \approx -0.9273 $$ 10. **Find $\theta$ values:** $$ \theta = 0.4115 - (-0.9273) = 1.3388 $$ $$ \theta = \pi - 0.4115 - (-0.9273) = 3.6574 $$ 11. **Final answer:** $$ \theta \approx 1.34 \text{ radians or } 3.66 \text{ radians (plus } 2k\pi \text{ for any integer } k) $$