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sin(x/2)<-(sqrt3/2)
Solving inequalities/ sin(x/2)<-(sqrt3/2)

sin(x/2)<-(sqrt3/2) inequation

A inequation with variable

The solution

You have entered [src] 
            ___ 
   /x\   -\/ 3  
sin|-| < -------
   \2/      2
sin⁡(x2)<−32\sin{\left(\frac{x}{2} \right)} < - \frac{\sqrt{3}}{2} 
sin(x/2) < -1*sqrt(3)/2
Detail solution
Given the inequality:
sin⁡(x2)<−32\sin{\left(\frac{x}{2} \right)} < - \frac{\sqrt{3}}{2}
To solve this inequality, we must first solve the corresponding equation:
sin⁡(x2)=−32\sin{\left(\frac{x}{2} \right)} = - \frac{\sqrt{3}}{2}
Solve:
Given the equation
sin⁡(x2)=−32\sin{\left(\frac{x}{2} \right)} = - \frac{\sqrt{3}}{2}
- this is the simplest trigonometric equation
This equation is transformed to
x2=2πn+asin⁡(−32)\frac{x}{2} = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}
x2=2πn−asin⁡(−32)+π\frac{x}{2} = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi
Or
x2=2πn−π3\frac{x}{2} = 2 \pi n - \frac{\pi}{3}
x2=2πn+4π3\frac{x}{2} = 2 \pi n + \frac{4 \pi}{3}
, where n - is a integer
Divide both parts of the equation by
12\frac{1}{2}
x1=4πn−2π3x_{1} = 4 \pi n - \frac{2 \pi}{3}
x2=4πn+8π3x_{2} = 4 \pi n + \frac{8 \pi}{3}
x1=4πn−2π3x_{1} = 4 \pi n - \frac{2 \pi}{3}
x2=4πn+8π3x_{2} = 4 \pi n + \frac{8 \pi}{3}
This roots
x1=4πn−2π3x_{1} = 4 \pi n - \frac{2 \pi}{3}
x2=4πn+8π3x_{2} = 4 \pi n + \frac{8 \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}
=
(4πn−2π3)−110\left(4 \pi n - \frac{2 \pi}{3}\right) - \frac{1}{10}
=
4πn−2π3−1104 \pi n - \frac{2 \pi}{3} - \frac{1}{10}
substitute to the expression
sin⁡(x2)<−32\sin{\left(\frac{x}{2} \right)} < - \frac{\sqrt{3}}{2}
sin⁡(4πn−2π3−1102)<−32\sin{\left(\frac{4 \pi n - \frac{2 \pi}{3} - \frac{1}{10}}{2} \right)} < - \frac{\sqrt{3}}{2}
                   ___ 
    /1    pi\   -\/ 3  
-sin|-- + --| < -------
    \20   3 /      2

one of the solutions of our inequality is:
x<4πn−2π3x < 4 \pi n - \frac{2 \pi}{3}
 _____           _____          
      \         /
-------ο-------ο-------
       x_1      x_2

Other solutions will get with the changeover to the next point
etc.
The answer:
x<4πn−2π3x < 4 \pi n - \frac{2 \pi}{3}
x>4πn+8π3x > 4 \pi n + \frac{8 \pi}{3}
Solving inequality on a graph
05-20-15-10-510152-2
Rapid solution [src] 
   /8*pi          10*pi\
And|---- < x, x < -----|
   \ 3              3  /
8π3<x∧x<10π3\frac{8 \pi}{3} < x \wedge x < \frac{10 \pi}{3} 
(8*pi/3 < x)∧(x < 10*pi/3)
Rapid solution 2 [src] 
 8*pi  10*pi 
(----, -----)
  3      3
x in (8π3,10π3)x\ in\ \left(\frac{8 \pi}{3}, \frac{10 \pi}{3}\right) 
x in Interval.open(8*pi/3, 10*pi/3)
The graph
sin(x/2)<-(sqrt3/2) inequation
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