Given the inequality:
$$- 2 \cos{\left(3 x \right)} + \sqrt{3} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- 2 \cos{\left(3 x \right)} + \sqrt{3} = 0$$
Solve:
Given the equation
$$- 2 \cos{\left(3 x \right)} + \sqrt{3} = 0$$
- this is the simplest trigonometric equation
Move sqrt(3) to right part of the equation
with the change of sign in sqrt(3)
We get:
$$- 2 \cos{\left(3 x \right)} = - \sqrt{3}$$
Divide both parts of the equation by -2
The equation is transformed to
$$\cos{\left(3 x \right)} = \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$3 x = \pi n + \operatorname{acos}{\left(\frac{\sqrt{3}}{2} \right)}$$
$$3 x = \pi n - \pi + \operatorname{acos}{\left(\frac{\sqrt{3}}{2} \right)}$$
Or
$$3 x = \pi n + \frac{\pi}{6}$$
$$3 x = \pi n - \frac{5 \pi}{6}$$
, where n - is a integer
Divide both parts of the equation by
$$3$$
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} - \frac{5 \pi}{18}$$
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} - \frac{5 \pi}{18}$$
This roots
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} - \frac{5 \pi}{18}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{3} + \frac{\pi}{18}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{3} - \frac{1}{10} + \frac{\pi}{18}$$
substitute to the expression
$$- 2 \cos{\left(3 x \right)} + \sqrt{3} < 0$$
$$- 2 \cos{\left(3 \left(\frac{\pi n}{3} - \frac{1}{10} + \frac{\pi}{18}\right) \right)} + \sqrt{3} < 0$$
___ / 3 pi \
\/ 3 - 2*cos|- -- + -- + pi*n| < 0
\ 10 6 / but
___ / 3 pi \
\/ 3 - 2*cos|- -- + -- + pi*n| > 0
\ 10 6 / Then
$$x < \frac{\pi n}{3} + \frac{\pi}{18}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{\pi n}{3} + \frac{\pi}{18} \wedge x < \frac{\pi n}{3} - \frac{5 \pi}{18}$$
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