1. **State the problem:** Solve the equation $$3 \tan x = 1 + 4 \cot x$$ for $$-180^\circ < x < 180^\circ$$. 2. **Recall definitions and formulas:** Recall that $$\cot x = \frac{1}{\tan x}$$. 3. **Rewrite the equation using $$\cot x = \frac{1}{\tan x}$$:** $$3 \tan x = 1 + 4 \frac{1}{\tan x}$$ 4. **Multiply both sides by $$\tan x$$ to clear the denominator:** $$3 \tan^2 x = \tan x + 4$$ 5. **Bring all terms to one side to form a quadratic in $$\tan x$$:** $$3 \tan^2 x - \tan x - 4 = 0$$ 6. **Use the quadratic formula for $$t = \tan x$$:** $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{1 \pm \sqrt{1 + 48}}{6} = \frac{1 \pm \sqrt{49}}{6}$$ 7. **Simplify the square root:** $$t = \frac{1 \pm 7}{6}$$ 8. **Find the two solutions for $$t$$:** - $$t_1 = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3}$$ - $$t_2 = \frac{1 - 7}{6} = \frac{-6}{6} = -1$$ 9. **Solve for $$x$$ using $$\tan x = t$$:** - For $$\tan x = \frac{4}{3}$$, find $$x$$ in $$(-180^\circ, 180^\circ)$$: $$x = \arctan\left(\frac{4}{3}\right) \approx 53.13^\circ$$ The tangent function has period $$180^\circ$$, so the other solution is: $$x = 53.13^\circ - 180^\circ = -126.87^\circ$$ - For $$\tan x = -1$$, find $$x$$ in $$(-180^\circ, 180^\circ)$$: $$x = \arctan(-1) = -45^\circ$$ The other solution is: $$x = -45^\circ + 180^\circ = 135^\circ$$ 10. **Final solutions rounded to nearest degree:** $$x \approx 53^\circ, -127^\circ, -45^\circ, 135^\circ$$ These are the roots of the equation in the given domain.