sin(x+π/6)-sin(x-π/6)=1/2

sinA –Sin B=2cos[ (A+B)/2]*sin[(A-B)/2]

Putting the above equation in identity:

sin(x+π/6)-sin(x-π/6)=2cos[(x+π/6)+(x-π/6)/2]*sin[(x+π/6)-(x-π/6)/2]

=2cos(x)*sin(π/6)=1/2

=2*cos(x)*1/2=1/2

cos x=1/2

x=π/3 , 4π/3

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