1. **State the problem:** Solve the trigonometric equation $3 + 2\cos^2(\theta)\tan(\theta) = 4\cos^2(\theta)$.\n\n2. **Recall the definitions and formulas:** \n- $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$\n- We will substitute $\tan(\theta)$ and simplify the equation.\n\n3. **Rewrite the equation using $\tan(\theta)$:**\n$$3 + 2\cos^2(\theta) \cdot \frac{\sin(\theta)}{\cos(\theta)} = 4\cos^2(\theta)$$\n\n4. **Simplify the left side:**\n$$3 + 2\cos(\theta)\sin(\theta) = 4\cos^2(\theta)$$\n\n5. **Bring all terms to one side:**\n$$3 + 2\cos(\theta)\sin(\theta) - 4\cos^2(\theta) = 0$$\n\n6. **Rewrite the equation:**\n$$2\cos(\theta)\sin(\theta) - 4\cos^2(\theta) = -3$$\n\n7. **Divide both sides by 2 to simplify:**\n$$\cos(\theta)\sin(\theta) - 2\cos^2(\theta) = -\frac{3}{2}$$\n\n8. **Use the double angle identity for sine:**\n$$\sin(2\theta) = 2\sin(\theta)\cos(\theta) \Rightarrow \sin(\theta)\cos(\theta) = \frac{\sin(2\theta)}{2}$$\n\n9. **Substitute into the equation:**\n$$\frac{\sin(2\theta)}{2} - 2\cos^2(\theta) = -\frac{3}{2}$$\n\n10. **Multiply both sides by 2:**\n$$\sin(2\theta) - 4\cos^2(\theta) = -3$$\n\n11. **Rewrite $\cos^2(\theta)$ using double angle identity:**\n$$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$\n\n12. **Substitute:**\n$$\sin(2\theta) - 4 \cdot \frac{1 + \cos(2\theta)}{2} = -3$$\n\n13. **Simplify:**\n$$\sin(2\theta) - 2(1 + \cos(2\theta)) = -3$$\n$$\sin(2\theta) - 2 - 2\cos(2\theta) = -3$$\n\n14. **Bring constants to one side:**\n$$\sin(2\theta) - 2\cos(2\theta) = -3 + 2 = -1$$\n\n15. **Rewrite:**\n$$\sin(2\theta) - 2\cos(2\theta) = -1$$\n\n16. **Let $x = 2\theta$, then:**\n$$\sin x - 2\cos x = -1$$\n\n17. **Express as a single sine function:**\nUse the identity $a\sin x + b\cos x = R\sin(x + \alpha)$ where $R = \sqrt{a^2 + b^2}$ and $\alpha = \arctan(\frac{b}{a})$. Here, $a=1$, $b=-2$.\n\n18. **Calculate $R$ and $\alpha$:**\n$$R = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$\n$$\alpha = \arctan\left(\frac{-2}{1}\right) = \arctan(-2)$$\n\n19. **Rewrite equation:**\n$$\sin x - 2\cos x = \sqrt{5} \sin(x + \alpha) = -1$$\n\n20. **Divide both sides by $\sqrt{5}$:**\n$$\sin(x + \alpha) = -\frac{1}{\sqrt{5}}$$\n\n21. **Solve for $x + \alpha$:**\n$$x + \alpha = \arcsin\left(-\frac{1}{\sqrt{5}}\right) + 2k\pi \quad \text{or} \quad x + \alpha = \pi - \arcsin\left(-\frac{1}{\sqrt{5}}\right) + 2k\pi$$\n\n22. **Calculate $\arcsin\left(-\frac{1}{\sqrt{5}}\right)$:**\nApproximately $-0.4636$ radians.\n\n23. **Write general solutions:**\n$$x + \alpha = -0.4636 + 2k\pi \quad \text{or} \quad x + \alpha = \pi + 0.4636 + 2k\pi$$\n\n24. **Recall $\alpha = \arctan(-2) \approx -1.1071$ radians, so:**\n$$x = 2\theta = -0.4636 - \alpha + 2k\pi = -0.4636 + 1.1071 + 2k\pi = 0.6435 + 2k\pi$$\n$$\text{or}$$\n$$2\theta = \pi + 0.4636 - \alpha + 2k\pi = 3.6052 + 2k\pi$$\n\n25. **Divide by 2 to find $\theta$:**\n$$\theta = 0.3217 + k\pi \quad \text{or} \quad \theta = 1.8026 + k\pi$$\n\n**Final answer:**\n$$\boxed{\theta = 0.3217 + k\pi \quad \text{or} \quad \theta = 1.8026 + k\pi, \quad k \in \mathbb{Z}}$$