√3 ctg x ≥ -1

To solve the inequality \(\sqrt{3} \cot x \geq -1\), we can first rewrite this in terms of \(\cot x\): \[ \cot x \geq -\frac{1}{\sqrt{3}} \] The cotangent function, \(\cot x\), is defined as \(\cot x = \frac{\cos x}{\sin x}\). The value \(-\frac{1}{\sqrt{3}}\) corresponds to a specific angle. First, think of the angles for which \(\cot x = -\frac{1}{\sqrt{3}}\): 1. \(\cot x = -\frac{1}{\sqrt{3}}\) occurs in the second quadrant and the fourth quadrant. 2. The angles where \(\cot x = \frac{1}{\sqrt{3}}\) are \(\frac{\pi}{6}\) (30 degrees) in the first quadrant and \( \frac{5\pi}{6}\) (150 degrees) in the second quadrant. Therefore, - \(\cot x = -\frac{1}{\sqrt{3}}\) happens at: - \( x = \frac{5\pi}{6} + k\pi \) (for \( k \in \mathbb{Z} \), in the second and fourth quadrants). Next, we need to find where the inequality holds. We know from the unit circle: - In the second quadrant (where \(x\) is between \(\frac{\pi}{2}\) and \(\pi\)), \(\cot x\) decreases from 0 to \(-\infty\). - In the fourth quadrant (where \(x\) is between \(3\pi/2\) and \(2\pi\)), \(\cot x\) increases from \(-\infty\) to 0. This means that \(\cot x\) is greater than or equal to \(-\frac{1}{\sqrt{3}}\): 1. In the second quadrant: \[ x \in \left[\frac{5\pi}{6}, \pi\right) \] 2. In the fourth quadrant: \[ x \in \left(0, \frac{\pi}{6}\right) + k\pi \] Specifically, the fourth quadrant gives: \[ x \in \left(3\frac{\pi}{2}, \frac{11\pi}{6}\right) \] Combining this information, the solution for the inequality \(\sqrt{3} \cot x \geq -1\) is given by: \[ x \in \left[\frac{5\pi}{6}, \pi\right) \cup \left(0, \frac{\pi}{6}\right) + k\pi, \quad k \in \mathbb{Z} \] Make sure to also consider the periods of the cotangent function when evaluating the specific intervals based on the angle ranges.