How do I solve sin(5x-pi/3)<=0, [-pi/2,pi/2]?

1 Answer
Apr 8, 2018

Given: sin(5x-pi/3)<=0, [-pi/2,pi/2]

First, do this without the domain restriction.

sin(5x-pi/3)<=0

This is true when the primary values of the argument are:

-pi <= 5x-pi/3<=0

Add pi3 to all 3:

-(2pi)/3 <= 5x<=pi/3

Add integer multiples of 2pi

-(2pi)/3+2pin <= 5x<=pi/3+2pin, n in ZZ

Divide everything by 5:

-(2pi)/15+2/5pin <= x<=pi/15+2/5pin, n in ZZ

I have graphed the domain restriction: color(blue)(-pi/2<= x <= pi/2) simultaneously with color(red)(-(2pi)/15+2/5pin <= x<=pi/15+2/5pin, n in ZZ):

www.desmos.com/calculatorwww.desmos.com/calculator

Please observe that there are 3 values of n, -1, 0 and 1, that fall within the domain but the leftmost (n = -1) will be truncated by the domain's lower bound:

n = -1:

-pi/2 <= x <= pi/15-2/5pi

n= 0

-(2pi)/15 <= x <= pi/15

n = 1

-(2pi)/15+2/5pi <= x<=pi/15+2/5pi

The union of the 3 domain segments is the solution to the inequality with the domain restriction.