How do you find all solutions of the equation #sin(x+pi/3)+sin(x-pi/3)=1# in the interval #[0,2pi)#?

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Answer 1 · 2017-08-10T01:49:01

#x = pi/2#

Use the following identities to solve this equation

#sin(A + B) = sinAcosB + sinBcosA#
#sin(A - B) = sinAcosB - sinBcosA#

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#sinxcos(pi/3) + sin(pi/3)cosx + sinxcos(pi/3) - sin(pi/3)cosx = 1#

#1/2sinx + sqrt(3)/2cosx + 1/2sinx - sqrt(3)/2cosx = 1#

The cosines obviously cancel each other out, so we are left with:

#1/2sinx + 1/2sinx = 1#

#sinx = 1#

#x = arcsin(1)#

#x= pi/2#

This will be our only solution on #[0, 2pi)#.

Hopefully this helps!