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Solving inequalities/ sin2x<(\sqrt(3))/(2)

sin2x<(\sqrt(3))/(2) inequation

A inequation with variable

The solution

You have entered [src] 
             ___
           \/ 3 
sin(2*x) < -----
             2
sin⁡(2x)<32\sin{\left(2 x \right)} < \frac{\sqrt{3}}{2} 
sin(2*x) < sqrt(3)/2
Detail solution
Given the inequality:
sin⁡(2x)<32\sin{\left(2 x \right)} < \frac{\sqrt{3}}{2}
To solve this inequality, we must first solve the corresponding equation:
sin⁡(2x)=32\sin{\left(2 x \right)} = \frac{\sqrt{3}}{2}
Solve:
Given the equation
sin⁡(2x)=32\sin{\left(2 x \right)} = \frac{\sqrt{3}}{2}
- this is the simplest trigonometric equation
This equation is transformed to
2x=2πn+asin⁡(32)2 x = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)}
2x=2πn−asin⁡(32)+π2 x = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)} + \pi
Or
2x=2πn+π32 x = 2 \pi n + \frac{\pi}{3}
2x=2πn+2π32 x = 2 \pi n + \frac{2 \pi}{3}
, where n - is a integer
Divide both parts of the equation by
22
x1=πn+π6x_{1} = \pi n + \frac{\pi}{6}
x2=πn+π3x_{2} = \pi n + \frac{\pi}{3}
x1=πn+π6x_{1} = \pi n + \frac{\pi}{6}
x2=πn+π3x_{2} = \pi n + \frac{\pi}{3}
This roots
x1=πn+π6x_{1} = \pi n + \frac{\pi}{6}
x2=πn+π3x_{2} = \pi n + \frac{\pi}{3}
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
x0<x1x_{0} < x_{1}
For example, let's take the point
x0=x1−110x_{0} = x_{1} - \frac{1}{10}
=
(πn+π6)+−110\left(\pi n + \frac{\pi}{6}\right) + - \frac{1}{10}
=
πn−110+π6\pi n - \frac{1}{10} + \frac{\pi}{6}
substitute to the expression
sin⁡(2x)<32\sin{\left(2 x \right)} < \frac{\sqrt{3}}{2}
sin⁡(2(πn−110+π6))<32\sin{\left(2 \left(\pi n - \frac{1}{10} + \frac{\pi}{6}\right) \right)} < \frac{\sqrt{3}}{2}
                           ___
   /  1   pi         \   \/ 3 
sin|- - + -- + 2*pi*n| < -----
   \  5   3          /     2

one of the solutions of our inequality is:
x<πn+π6x < \pi n + \frac{\pi}{6}
 _____           _____          
      \         /
-------ο-------ο-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
x<πn+π6x < \pi n + \frac{\pi}{6}
x>πn+π3x > \pi n + \frac{\pi}{3}
Solving inequality on a graph
0-70-60-50-40-30-20-10102030405060702-2
Rapid solution 2 [src] 
    pi     pi     
[0, --) U (--, pi]
    6      3
x in [0,π6)∪(π3,π]x\ in\ \left[0, \frac{\pi}{6}\right) \cup \left(\frac{\pi}{3}, \pi\right] 
x in Union(Interval.Ropen(0, pi/6), Interval.Lopen(pi/3, pi))
Rapid solution [src] 
  /   /            pi\     /         pi    \\
Or|And|0 <= x, x < --|, And|x <= pi, -- < x||
  \   \            6 /     \         3     //
(0≤x∧x<π6)∨(x≤π∧π3<x)\left(0 \leq x \wedge x < \frac{\pi}{6}\right) \vee \left(x \leq \pi \wedge \frac{\pi}{3} < x\right) 
((0 <= x)∧(x < pi/6))∨((x <= pi)∧(pi/3 < x))
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