1. **State the problem:** Solve the equation $$\tan x + 1 = -\sqrt{3} - \sqrt{3} \cot x$$ for $$x$$ in the interval $$[0, 2\pi)$$. 2. **Rewrite the equation:** Recall that $$\cot x = \frac{1}{\tan x}$$, so substitute to get: $$\tan x + 1 = -\sqrt{3} - \sqrt{3} \cdot \frac{1}{\tan x}$$ 3. **Multiply both sides by $$\tan x$$ to clear the denominator:** $$\tan x \cdot \tan x + 1 \cdot \tan x = -\sqrt{3} \tan x - \sqrt{3}$$ which is $$\tan^2 x + \tan x = -\sqrt{3} \tan x - \sqrt{3}$$ 4. **Bring all terms to one side:** $$\tan^2 x + \tan x + \sqrt{3} \tan x + \sqrt{3} = 0$$ 5. **Combine like terms:** $$\tan^2 x + (1 + \sqrt{3}) \tan x + \sqrt{3} = 0$$ 6. **Let $$t = \tan x$$ and solve the quadratic:** $$t^2 + (1 + \sqrt{3}) t + \sqrt{3} = 0$$ 7. **Use the quadratic formula:** $$t = \frac{-(1 + \sqrt{3}) \pm \sqrt{(1 + \sqrt{3})^2 - 4 \cdot 1 \cdot \sqrt{3}}}{2}$$ 8. **Calculate the discriminant:** $$(1 + \sqrt{3})^2 - 4 \sqrt{3} = (1 + 2\sqrt{3} + 3) - 4\sqrt{3} = 4 + 2\sqrt{3} - 4\sqrt{3} = 4 - 2\sqrt{3}$$ 9. **Simplify the square root:** $$\sqrt{4 - 2\sqrt{3}}$$ Note that $$4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$$, so $$\sqrt{4 - 2\sqrt{3}} = \sqrt{3} - 1$$ 10. **Substitute back:** $$t = \frac{-(1 + \sqrt{3}) \pm (\sqrt{3} - 1)}{2}$$ 11. **Calculate the two roots:** - For the plus sign: $$t = \frac{-(1 + \sqrt{3}) + (\sqrt{3} - 1)}{2} = \frac{-1 - \sqrt{3} + \sqrt{3} - 1}{2} = \frac{-2}{2} = -1$$ - For the minus sign: $$t = \frac{-(1 + \sqrt{3}) - (\sqrt{3} - 1)}{2} = \frac{-1 - \sqrt{3} - \sqrt{3} + 1}{2} = \frac{-2\sqrt{3}}{2} = -\sqrt{3}$$ 12. **Solve for $$x$$:** - When $$\tan x = -1$$, $$x = \frac{3\pi}{4}, \frac{7\pi}{4}$$ in $$[0, 2\pi)$$. - When $$\tan x = -\sqrt{3}$$, $$x = \frac{5\pi}{6}, \frac{11\pi}{6}$$ in $$[0, 2\pi)$$. **Final answer:** $$x = \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{5\pi}{6}, \frac{11\pi}{6}$$