Given the inequality:
$$\cos{\left(3 x \right)} > \frac{\sqrt{2}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(3 x \right)} = \frac{\sqrt{2}}{2}$$
Solve:
Given the equation
$$\cos{\left(3 x \right)} = \frac{\sqrt{2}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$3 x = \pi n + \operatorname{acos}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$3 x = \pi n - \pi + \operatorname{acos}{\left(\frac{\sqrt{2}}{2} \right)}$$
Or
$$3 x = \pi n + \frac{\pi}{4}$$
$$3 x = \pi n - \frac{3 \pi}{4}$$
, where n - is a integer
Divide both parts of the equation by
$$3$$
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{12}$$
$$x_{2} = \frac{\pi n}{3} - \frac{\pi}{4}$$
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{12}$$
$$x_{2} = \frac{\pi n}{3} - \frac{\pi}{4}$$
This roots
$$x_{1} = \frac{\pi n}{3} + \frac{\pi}{12}$$
$$x_{2} = \frac{\pi n}{3} - \frac{\pi}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{3} + \frac{\pi}{12}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{3} - \frac{1}{10} + \frac{\pi}{12}$$
substitute to the expression
$$\cos{\left(3 x \right)} > \frac{\sqrt{2}}{2}$$
$$\cos{\left(3 \left(\frac{\pi n}{3} - \frac{1}{10} + \frac{\pi}{12}\right) \right)} > \frac{\sqrt{2}}{2}$$
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/ 3 pi \ \/ 2
cos|- -- + -- + pi*n| > -----
\ 10 4 / 2
Then
$$x < \frac{\pi n}{3} + \frac{\pi}{12}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{\pi n}{3} + \frac{\pi}{12} \wedge x < \frac{\pi n}{3} - \frac{\pi}{4}$$
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