ctg6x>-sqrt3 inequation

Given the inequality:
$$\cot{\left(6 x \right)} > - \sqrt{3}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot{\left(6 x \right)} = - \sqrt{3}$$
Solve:
Given the equation
$$\cot{\left(6 x \right)} = - \sqrt{3}$$
transform
$$\cot{\left(6 x \right)} + \sqrt{3} = 0$$
$$\cot{\left(6 x \right)} + \sqrt{3} = 0$$
Do replacement
$$w = \cot{\left(6 x \right)}$$
Expand brackets in the left part

w + sqrt3 = 0

Divide both parts of the equation by (w + sqrt(3))/w

w = 0 / ((w + sqrt(3))/w)

We get the answer: w = -sqrt(3)
do backward replacement
$$\cot{\left(6 x \right)} = w$$
substitute w:
$$x_{1} = - \frac{\pi}{36}$$
$$x_{1} = - \frac{\pi}{36}$$
This roots
$$x_{1} = - \frac{\pi}{36}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} - \frac{\pi}{36}$$
=
$$- \frac{1}{10} - \frac{\pi}{36}$$
substitute to the expression
$$\cot{\left(6 x \right)} > - \sqrt{3}$$
$$\cot{\left(6 \left(- \frac{1}{10} - \frac{\pi}{36}\right) \right)} > - \sqrt{3}$$

    /3   pi\      ___
-cot|- + --| > -\/ 3 
    \5   6 /   

the solution of our inequality is:
$$x < - \frac{\pi}{36}$$

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