How do you solve #4sin(3x)=2# in the interval #[0,(4pi)/3]#?

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Answer 1 · 2017-02-02T22:35:11

The solutions included in the interval #[0;4/3pi]# are

#pi/18; 5/18pi; 13/18pi; 17/18pi#

It is:

#sin(3x)=2/4=1/2#

then

#3x=pi/6+2kpi# or #3x=pi-pi/6+2kpi#

#x=pi/18+2/3kpi# or #x=5/18pi+2/3kpi#

The solutions included in the interval #[0;4/3pi]# are therefore:

#pi/18; 5/18pi; 13/18pi; 17/18pi#