How do you find all solutions of the equation #sin(x+pi/6)-sin(x-pi/6)=1/2# in the interval #[0,2pi)#?

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Answer 1 · 2017-10-26T06:25:01

#x=pi/6# or #(11pi)/6#

We have the identity #sin(A+B)-sin(A-B)=2cosAsinB#

hence #sin(x+pi/6)-sin(x-pi/6)=1/2# can be written as

#2cosxsin(pi/6)=1/2#

or #2cosx xx1/2=1/2#

or #cosx=1/2=cos(pi/6)#

and in the interval #[0,2pi)#

#x=pi/6# or #2pi-pi/6=(11pi)/6#