1. **State the problem:** Solve the equation $$4 \cos(x - 2) - 3 \cos(x - 1) = 2 \cos 6$$ for $x$. 2. **Recall the cosine subtraction formula:** $$\cos(a - b) = \cos a \cos b + \sin a \sin b$$ 3. **Apply the formula to each cosine term:** $$4 \cos(x - 2) = 4(\cos x \cos 2 + \sin x \sin 2)$$ $$-3 \cos(x - 1) = -3(\cos x \cos 1 + \sin x \sin 1)$$ 4. **Rewrite the left side:** $$4 \cos x \cos 2 + 4 \sin x \sin 2 - 3 \cos x \cos 1 - 3 \sin x \sin 1$$ 5. **Group like terms:** $$\cos x (4 \cos 2 - 3 \cos 1) + \sin x (4 \sin 2 - 3 \sin 1) = 2 \cos 6$$ 6. **Calculate constants (approximate):** $$4 \cos 2 \approx 4 \times (-0.4161) = -1.6644$$ $$3 \cos 1 \approx 3 \times 0.5403 = 1.6209$$ $$4 \sin 2 \approx 4 \times 0.9093 = 3.6372$$ $$3 \sin 1 \approx 3 \times 0.8415 = 2.5245$$ 7. **Substitute constants:** $$\cos x (-1.6644 - 1.6209) + \sin x (3.6372 - 2.5245) = 2 \cos 6$$ $$\cos x (-3.2853) + \sin x (1.1127) = 2 \cos 6$$ 8. **Calculate right side:** $$2 \cos 6 \approx 2 \times 0.9602 = 1.9204$$ 9. **Rewrite equation:** $$-3.2853 \cos x + 1.1127 \sin x = 1.9204$$ 10. **Divide entire equation by $\sqrt{(-3.2853)^2 + (1.1127)^2}$ to normalize coefficients:** $$\sqrt{(-3.2853)^2 + (1.1127)^2} = \sqrt{10.785 + 1.238} = \sqrt{12.023} \approx 3.469$$ $$\frac{-3.2853}{3.469} \cos x + \frac{1.1127}{3.469} \sin x = \frac{1.9204}{3.469}$$ 11. **Simplify:** $$-0.947 \cos x + 0.321 \sin x = 0.553$$ 12. **Rewrite as single cosine using phase shift:** $$R \cos(x + \alpha) = 0.553$$ where $$R = \sqrt{(-0.947)^2 + (0.321)^2} = 1$$ and $$\tan \alpha = \frac{0.321}{-0.947} = -0.339$$ 13. **Calculate $\alpha$:** $$\alpha = \arctan(-0.339) \approx -0.327 \text{ radians}$$ 14. **Rewrite equation:** $$\cos(x - 0.327) = 0.553$$ 15. **Solve for $x$:** $$x - 0.327 = \pm \arccos(0.553) + 2k\pi$$ $$x = 0.327 \pm 0.988 + 2k\pi$$ 16. **Final solutions:** $$x = 1.315 + 2k\pi \quad \text{or} \quad x = -0.661 + 2k\pi, \quad k \in \mathbb{Z}$$