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Solving inequalities/ sin<=-(√3)/2

sin<=-(√3)/2 inequation

A inequation with variable

The solution

You have entered [src] 
             ___ 
          -\/ 3  
sin(x) <= -------
             2
sin⁡(x)≤(−1)32\sin{\left(x \right)} \leq \frac{\left(-1\right) \sqrt{3}}{2} 
sin(x) <= (-sqrt(3))/2
Detail solution
Given the inequality:
sin⁡(x)≤(−1)32\sin{\left(x \right)} \leq \frac{\left(-1\right) \sqrt{3}}{2}
To solve this inequality, we must first solve the corresponding equation:
sin⁡(x)=(−1)32\sin{\left(x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}
Solve:
Given the equation
sin⁡(x)=(−1)32\sin{\left(x \right)} = \frac{\left(-1\right) \sqrt{3}}{2}
- this is the simplest trigonometric equation
This equation is transformed to
x=2πn+asin⁡(−32)x = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}
x=2πn−asin⁡(−32)+πx = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi
Or
x=2πn−π3x = 2 \pi n - \frac{\pi}{3}
x=2πn+4π3x = 2 \pi n + \frac{4 \pi}{3}
, where n - is a integer
x1=2πn−π3x_{1} = 2 \pi n - \frac{\pi}{3}
x2=2πn+4π3x_{2} = 2 \pi n + \frac{4 \pi}{3}
x1=2πn−π3x_{1} = 2 \pi n - \frac{\pi}{3}
x2=2πn+4π3x_{2} = 2 \pi n + \frac{4 \pi}{3}
This roots
x1=2πn−π3x_{1} = 2 \pi n - \frac{\pi}{3}
x2=2πn+4π3x_{2} = 2 \pi n + \frac{4 \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} \leq x_{1}
For example, let's take the point
x0=x1−110x_{0} = x_{1} - \frac{1}{10}
=
(2πn−π3)+−110\left(2 \pi n - \frac{\pi}{3}\right) + - \frac{1}{10}
=
2πn−π3−1102 \pi n - \frac{\pi}{3} - \frac{1}{10}
substitute to the expression
sin⁡(x)≤(−1)32\sin{\left(x \right)} \leq \frac{\left(-1\right) \sqrt{3}}{2}
sin⁡(2πn−π3−110)≤(−1)32\sin{\left(2 \pi n - \frac{\pi}{3} - \frac{1}{10} \right)} \leq \frac{\left(-1\right) \sqrt{3}}{2}
                             ___ 
    /1    pi         \    -\/ 3  
-sin|-- + -- - 2*pi*n| <= -------
    \10   3          /       2

one of the solutions of our inequality is:
x≤2πn−π3x \leq 2 \pi n - \frac{\pi}{3}
 _____           _____          
      \         /
-------•-------•-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
x≤2πn−π3x \leq 2 \pi n - \frac{\pi}{3}
x≥2πn+4π3x \geq 2 \pi n + \frac{4 \pi}{3}
Solving inequality on a graph
0-60-50-40-30-20-101020304050602-2
Rapid solution [src] 
   /4*pi            5*pi\
And|---- <= x, x <= ----|
   \ 3               3  /
4π3≤x∧x≤5π3\frac{4 \pi}{3} \leq x \wedge x \leq \frac{5 \pi}{3} 
(4*pi/3 <= x)∧(x <= 5*pi/3)
Rapid solution 2 [src] 
 4*pi  5*pi 
[----, ----]
  3     3
x in [4π3,5π3]x\ in\ \left[\frac{4 \pi}{3}, \frac{5 \pi}{3}\right] 
x in Interval(4*pi/3, 5*pi/3)
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