Mister Exam

Other calculators

Solving inequalities/ sin3x<0.5

sin3x<0.5 inequation

A inequation with variable

The solution

You have entered [src] 
sin(3*x) < 1/2
sin⁡(3x)<12\sin{\left(3 x \right)} < \frac{1}{2} 
sin(3*x) < 1/2
Detail solution
Given the inequality:
sin⁡(3x)<12\sin{\left(3 x \right)} < \frac{1}{2}
To solve this inequality, we must first solve the corresponding equation:
sin⁡(3x)=12\sin{\left(3 x \right)} = \frac{1}{2}
Solve:
Given the equation
sin⁡(3x)=12\sin{\left(3 x \right)} = \frac{1}{2}
- this is the simplest trigonometric equation
This equation is transformed to
3x=2πn+asin⁡(12)3 x = 2 \pi n + \operatorname{asin}{\left(\frac{1}{2} \right)}
3x=2πn−asin⁡(12)+π3 x = 2 \pi n - \operatorname{asin}{\left(\frac{1}{2} \right)} + \pi
Or
3x=2πn+π63 x = 2 \pi n + \frac{\pi}{6}
3x=2πn+5π63 x = 2 \pi n + \frac{5 \pi}{6}
, where n - is a integer
Divide both parts of the equation by
33
x1=2πn3+π18x_{1} = \frac{2 \pi n}{3} + \frac{\pi}{18}
x2=2πn3+5π18x_{2} = \frac{2 \pi n}{3} + \frac{5 \pi}{18}
x1=2πn3+π18x_{1} = \frac{2 \pi n}{3} + \frac{\pi}{18}
x2=2πn3+5π18x_{2} = \frac{2 \pi n}{3} + \frac{5 \pi}{18}
This roots
x1=2πn3+π18x_{1} = \frac{2 \pi n}{3} + \frac{\pi}{18}
x2=2πn3+5π18x_{2} = \frac{2 \pi n}{3} + \frac{5 \pi}{18}
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}
=
(2πn3+π18)+−110\left(\frac{2 \pi n}{3} + \frac{\pi}{18}\right) + - \frac{1}{10}
=
2πn3−110+π18\frac{2 \pi n}{3} - \frac{1}{10} + \frac{\pi}{18}
substitute to the expression
sin⁡(3x)<12\sin{\left(3 x \right)} < \frac{1}{2}
sin⁡(3(2πn3−110+π18))<12\sin{\left(3 \left(\frac{2 \pi n}{3} - \frac{1}{10} + \frac{\pi}{18}\right) \right)} < \frac{1}{2}
   /  3    pi         \      
sin|- -- + -- + 2*pi*n| < 1/2
   \  10   6          /

one of the solutions of our inequality is:
x<2πn3+π18x < \frac{2 \pi n}{3} + \frac{\pi}{18}
 _____           _____          
      \         /
-------ο-------ο-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
x<2πn3+π18x < \frac{2 \pi n}{3} + \frac{\pi}{18}
x>2πn3+5π18x > \frac{2 \pi n}{3} + \frac{5 \pi}{18}
Solving inequality on a graph
0-80-60-40-20204060802-2
Rapid solution 2 [src] 
    pi     5*pi  2*pi 
[0, --) U (----, ----]
    18      18    3
x in [0,π18)∪(5π18,2π3]x\ in\ \left[0, \frac{\pi}{18}\right) \cup \left(\frac{5 \pi}{18}, \frac{2 \pi}{3}\right] 
x in Union(Interval.Ropen(0, pi/18), Interval.Lopen(5*pi/18, 2*pi/3))
Rapid solution [src] 
  /   /            pi\     /     2*pi  5*pi    \\
Or|And|0 <= x, x < --|, And|x <= ----, ---- < x||
  \   \            18/     \      3     18     //
(0≤x∧x<π18)∨(x≤2π3∧5π18<x)\left(0 \leq x \wedge x < \frac{\pi}{18}\right) \vee \left(x \leq \frac{2 \pi}{3} \wedge \frac{5 \pi}{18} < x\right) 
((0 <= x)∧(x < pi/18))∨((x <= 2*pi/3)∧(5*pi/18 < x))
    © Mister Exam – Calculator