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-2*cos(x/3)>1 inequation

A inequation with variable

The solution

You have entered [src]
      /x\    
-2*cos|-| > 1
      \3/    
$$- 2 \cos{\left(\frac{x}{3} \right)} > 1$$
-2*cos(x/3) > 1
Detail solution
Given the inequality:
$$- 2 \cos{\left(\frac{x}{3} \right)} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$- 2 \cos{\left(\frac{x}{3} \right)} = 1$$
Solve:
Given the equation
$$- 2 \cos{\left(\frac{x}{3} \right)} = 1$$
- this is the simplest trigonometric equation
Divide both parts of the equation by -2

The equation is transformed to
$$\cos{\left(\frac{x}{3} \right)} = - \frac{1}{2}$$
This equation is transformed to
$$\frac{x}{3} = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$\frac{x}{3} = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$\frac{x}{3} = \pi n + \frac{2 \pi}{3}$$
$$\frac{x}{3} = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{3}$$
$$x_{1} = 3 \pi n + 2 \pi$$
$$x_{2} = 3 \pi n - \pi$$
$$x_{1} = 3 \pi n + 2 \pi$$
$$x_{2} = 3 \pi n - \pi$$
This roots
$$x_{1} = 3 \pi n + 2 \pi$$
$$x_{2} = 3 \pi n - \pi$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(3 \pi n + 2 \pi\right) + - \frac{1}{10}$$
=
$$3 \pi n - \frac{1}{10} + 2 \pi$$
substitute to the expression
$$- 2 \cos{\left(\frac{x}{3} \right)} > 1$$
$$- 2 \cos{\left(\frac{3 \pi n - \frac{1}{10} + 2 \pi}{3} \right)} > 1$$
     /  1    pi       \    
2*sin|- -- + -- + pi*n| > 1
     \  30   6        /    

Then
$$x < 3 \pi n + 2 \pi$$
no execute
one of the solutions of our inequality is:
$$x > 3 \pi n + 2 \pi \wedge x < 3 \pi n - \pi$$
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        /     \  
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       x1      x2
Solving inequality on a graph
Rapid solution [src]
And(2*pi < x, x < 4*pi)
$$2 \pi < x \wedge x < 4 \pi$$
(2*pi < x)∧(x < 4*pi)
Rapid solution 2 [src]
(2*pi, 4*pi)
$$x\ in\ \left(2 \pi, 4 \pi\right)$$
x in Interval.open(2*pi, 4*pi)