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sin(3x)≥√2/2 inequation

A inequation with variable

The solution

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            \/ 2 
sin(3*x) >= -----
              2  
sin(3x)22\sin{\left(3 x \right)} \geq \frac{\sqrt{2}}{2}
sin(3*x) >= sqrt(2)/2
Detail solution
Given the inequality:
sin(3x)22\sin{\left(3 x \right)} \geq \frac{\sqrt{2}}{2}
To solve this inequality, we must first solve the corresponding equation:
sin(3x)=22\sin{\left(3 x \right)} = \frac{\sqrt{2}}{2}
Solve:
Given the equation
sin(3x)=22\sin{\left(3 x \right)} = \frac{\sqrt{2}}{2}
- this is the simplest trigonometric equation
This equation is transformed to
3x=2πn+asin(22)3 x = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)}
3x=2πnasin(22)+π3 x = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)} + \pi
Or
3x=2πn+π43 x = 2 \pi n + \frac{\pi}{4}
3x=2πn+3π43 x = 2 \pi n + \frac{3 \pi}{4}
, where n - is a integer
Divide both parts of the equation by
33
x1=2πn3+π12x_{1} = \frac{2 \pi n}{3} + \frac{\pi}{12}
x2=2πn3+π4x_{2} = \frac{2 \pi n}{3} + \frac{\pi}{4}
x1=2πn3+π12x_{1} = \frac{2 \pi n}{3} + \frac{\pi}{12}
x2=2πn3+π4x_{2} = \frac{2 \pi n}{3} + \frac{\pi}{4}
This roots
x1=2πn3+π12x_{1} = \frac{2 \pi n}{3} + \frac{\pi}{12}
x2=2πn3+π4x_{2} = \frac{2 \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:
x0x1x_{0} \leq x_{1}
For example, let's take the point
x0=x1110x_{0} = x_{1} - \frac{1}{10}
=
(2πn3+π12)+110\left(\frac{2 \pi n}{3} + \frac{\pi}{12}\right) + - \frac{1}{10}
=
2πn3110+π12\frac{2 \pi n}{3} - \frac{1}{10} + \frac{\pi}{12}
substitute to the expression
sin(3x)22\sin{\left(3 x \right)} \geq \frac{\sqrt{2}}{2}
sin(3(2πn3110+π12))22\sin{\left(3 \left(\frac{2 \pi n}{3} - \frac{1}{10} + \frac{\pi}{12}\right) \right)} \geq \frac{\sqrt{2}}{2}
                             ___
   /  3    pi         \    \/ 2 
sin|- -- + -- + 2*pi*n| >= -----
   \  10   4          /      2  
                           

but
                            ___
   /  3    pi         \   \/ 2 
sin|- -- + -- + 2*pi*n| < -----
   \  10   4          /     2  
                          

Then
x2πn3+π12x \leq \frac{2 \pi n}{3} + \frac{\pi}{12}
no execute
one of the solutions of our inequality is:
x2πn3+π12x2πn3+π4x \geq \frac{2 \pi n}{3} + \frac{\pi}{12} \wedge x \leq \frac{2 \pi n}{3} + \frac{\pi}{4}
         _____  
        /     \  
-------•-------•-------
       x1      x2
Solving inequality on a graph
0-80-60-40-20204060802-2
Rapid solution [src]
   /pi            pi\
And|-- <= x, x <= --|
   \12            4 /
π12xxπ4\frac{\pi}{12} \leq x \wedge x \leq \frac{\pi}{4}
(pi/12 <= x)∧(x <= pi/4)
Rapid solution 2 [src]
 pi  pi 
[--, --]
 12  4  
x in [π12,π4]x\ in\ \left[\frac{\pi}{12}, \frac{\pi}{4}\right]
x in Interval(pi/12, pi/4)