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.

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