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

1 Answer
Douglas K.
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/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.

Related questions
Impact of this question
1757 views around the world
You can reuse this answer
Creative Commons License