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Solving inequalities/ sint<0

sint<0 inequation

A inequation with variable

The solution

You have entered [src] 
sin(t) < 0
sin⁡(t)<0\sin{\left(t \right)} < 0 
sin(t) < 0
Detail solution
Given the inequality:
sin⁡(t)<0\sin{\left(t \right)} < 0
To solve this inequality, we must first solve the corresponding equation:
sin⁡(t)=0\sin{\left(t \right)} = 0
Solve:
Given the equation
sin⁡(t)=0\sin{\left(t \right)} = 0
- this is the simplest trigonometric equation
with the change of sign in 0

We get:
sin⁡(t)=0\sin{\left(t \right)} = 0
This equation is transformed to
t=2πn+asin⁡(0)t = 2 \pi n + \operatorname{asin}{\left(0 \right)}
t=2πn−asin⁡(0)+πt = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi
Or
t=2πnt = 2 \pi n
t=2πn+πt = 2 \pi n + \pi
, where n - is a integer
t1=2πnt_{1} = 2 \pi n
t2=2πn+πt_{2} = 2 \pi n + \pi
t1=2πnt_{1} = 2 \pi n
t2=2πn+πt_{2} = 2 \pi n + \pi
This roots
t1=2πnt_{1} = 2 \pi n
t2=2πn+πt_{2} = 2 \pi n + \pi
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
t0<t1t_{0} < t_{1}
For example, let's take the point
t0=t1−110t_{0} = t_{1} - \frac{1}{10}
=
2πn+−1102 \pi n + - \frac{1}{10}
=
2πn−1102 \pi n - \frac{1}{10}
substitute to the expression
sin⁡(t)<0\sin{\left(t \right)} < 0
sin⁡(2πn−110)<0\sin{\left(2 \pi n - \frac{1}{10} \right)} < 0
sin(-1/10 + 2*pi*n) < 0

one of the solutions of our inequality is:
t<2πnt < 2 \pi n
 _____           _____          
      \         /
-------ο-------ο-------
       t1      t2

Other solutions will get with the changeover to the next point
etc.
The answer:
t<2πnt < 2 \pi n
t>2πn+πt > 2 \pi n + \pi
Solving inequality on a graph
0-60-50-40-30-20-101020304050602-2
Rapid solution 2 [src] 
(pi, 2*pi)
t in (π,2π)t\ in\ \left(\pi, 2 \pi\right) 
t in Interval.open(pi, 2*pi)
Rapid solution [src] 
And(pi < t, t < 2*pi)
π<t∧t<2π\pi < t \wedge t < 2 \pi 
(pi < t)∧(t < 2*pi)
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