Given the inequality:
$$3 \sqrt{3} \tan{\left(x \right)} + 3 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$3 \sqrt{3} \tan{\left(x \right)} + 3 = 0$$
Solve:
Given the equation
$$3 \sqrt{3} \tan{\left(x \right)} + 3 = 0$$
- this is the simplest trigonometric equation
Move 3 to right part of the equation
with the change of sign in 3
We get:
$$3 \sqrt{3} \tan{\left(x \right)} = -3$$
Divide both parts of the equation by 3*sqrt(3)
The equation is transformed to
$$\tan{\left(x \right)} = - \frac{\sqrt{3}}{3}$$
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(- \frac{\sqrt{3}}{3} \right)}$$
Or
$$x = \pi n - \frac{\pi}{6}$$
, where n - is a integer
$$x_{1} = \pi n - \frac{\pi}{6}$$
$$x_{1} = \pi n - \frac{\pi}{6}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$3 \sqrt{3} \tan{\left(x \right)} + 3 < 0$$
$$3 \sqrt{3} \tan{\left(\pi n - \frac{\pi}{6} - \frac{1}{10} \right)} + 3 < 0$$
___ /1 pi \
3 - 3*\/ 3 *tan|-- + -- - pi*n| < 0
\10 6 / the solution of our inequality is:
$$x < \pi n - \frac{\pi}{6}$$
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