1. **State the problem:** Solve the trigonometric equation $4\sin x + 3\cos x = 2$ for $x$. 2. **Formula and approach:** We use the method of expressing $a\sin x + b\cos x$ as $R\sin(x+\alpha)$ where $R = \sqrt{a^2 + b^2}$ and $\alpha$ satisfies $\cos \alpha = \frac{a}{R}$, $\sin \alpha = \frac{b}{R}$. 3. Calculate $R$: $$R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ 4. Find $\alpha$: $$\cos \alpha = \frac{4}{5}, \quad \sin \alpha = \frac{3}{5}$$ 5. Rewrite the equation: $$4\sin x + 3\cos x = 5\sin(x + \alpha) = 2$$ 6. Solve for $\sin(x + \alpha)$: $$\sin(x + \alpha) = \frac{2}{5}$$ 7. Find general solutions for $x + \alpha$: $$x + \alpha = \sin^{-1}\left(\frac{2}{5}\right) + 2k\pi \quad \text{or} \quad x + \alpha = \pi - \sin^{-1}\left(\frac{2}{5}\right) + 2k\pi, \quad k \in \mathbb{Z}$$ 8. Substitute back $\alpha$: $$x = -\alpha + \sin^{-1}\left(\frac{2}{5}\right) + 2k\pi \quad \text{or} \quad x = -\alpha + \pi - \sin^{-1}\left(\frac{2}{5}\right) + 2k\pi$$ 9. This gives the complete solution set for $x$. **Final answer:** $$x = -\arccos\left(\frac{4}{5}\right) + \arcsin\left(\frac{2}{5}\right) + 2k\pi \quad \text{or} \quad x = -\arccos\left(\frac{4}{5}\right) + \pi - \arcsin\left(\frac{2}{5}\right) + 2k\pi, \quad k \in \mathbb{Z}$$