Given the inequality:
$$\cos{\left(6 x \right)} > - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(6 x \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(6 x \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$6 x = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$6 x = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$6 x = \pi n + \frac{2 \pi}{3}$$
$$6 x = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$6$$
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
This roots
$$x_{1} = \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{18}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{6} + \frac{\pi}{9}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{6} - \frac{1}{10} + \frac{\pi}{9}$$
substitute to the expression
$$\cos{\left(6 x \right)} > - \frac{1}{2}$$
$$\cos{\left(6 \left(\frac{\pi n}{6} - \frac{1}{10} + \frac{\pi}{9}\right) \right)} > - \frac{1}{2}$$
/ 3 pi \
-sin|- - + -- + pi*n| > -1/2
\ 5 6 /
one of the solutions of our inequality is:
$$x < \frac{\pi n}{6} + \frac{\pi}{9}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{6} + \frac{\pi}{9}$$
$$x > \frac{\pi n}{6} - \frac{\pi}{18}$$