To solve the given trigonometric equation:

$\sqrt{3} \cot x = -3$

for $0 \leq x < 2\pi$, let us proceed step by step.

Step 1: Isolate $\cot x$

Divide both sides of the equation by $\sqrt{3}$: $\cot x = -\frac{3}{\sqrt{3}} = -\sqrt{3}.$

Step 2: Recall the definition of $\cot x$

$\cot x = \frac{\cos x}{\sin x}.$ The equation implies that: $\frac{\cos x}{\sin x} = -\sqrt{3}.$

Step 3: Determine the reference angle

The reference angle is determined by considering the absolute value of $\cot x = \sqrt{3}$. From standard trigonometric values: $\cot \frac{\pi}{6} = \sqrt{3}.$ Thus, the reference angle is: $\alpha = \frac{\pi}{6}.$

Step 4: Find the quadrants where $\cot x < 0$

The cotangent function is negative in the second and fourth quadrants. Thus, the solutions are: $x = \pi - \alpha \quad \text{(second quadrant)}$ $x = 2\pi - \alpha \quad \text{(fourth quadrant)}.$

Step 5: Substitute the reference angle

$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6},$ $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}.$

Final Answer:

The solutions are: $x = \frac{5\pi}{6}, \frac{11\pi}{6}.$

Would you like further explanation or visualizations for any step?

Related Questions:

1. What is the general solution for $\cot x = -\sqrt{3}$?
2. How do you determine the reference angle for other trigonometric ratios?
3. What is the domain of the cotangent function?
4. Why is cotangent negative in the second and fourth quadrants?
5. Can this method be extended to solve similar equations, such as $\cot x = -2$?

Tip:

Always verify which quadrants satisfy the sign of the trigonometric function when solving equations!