Question #fab30

1 Answer
Martin C.
Feb 3, 2018

#x_1=n/2pi#
#or#
#x_2=n/3pi-pi/6#

Explanation:

To solve this matter we look at the products at their own. If one product equals zero the result will be zero as well.

#sin(2x)=0 or cos(3x)=0#
#sin(2x)=0 |sin^-1()#
#2x=sin(0)^-1#
#2x=npi |:2#
#x_1=n/2pi#
#or#
#cos(3x)=0|cos^-1()#
#3x=cos^-1(0)#
#3x=npi-pi/2|:3#
#x_2=n/3pi-pi/6#

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