Question 174444
{{{sin(x+pi/6)-sin(x-pi/6)=1/2}}} Start with the given equation



{{{(sin(x)cos(pi/6)+cos(x)sin(pi/6))-sin(x-pi/6)=1/2}}} Expand {{{sin(x+pi/6)}}}



{{{(sin(x)cos(pi/6)+cos(x)sin(pi/6))-(sin(x)cos(pi/6)-cos(x)sin(pi/6))=1/2}}} Expand {{{sin(x-pi/6)}}}



{{{sin(x)cos(pi/6)+cos(x)sin(pi/6)-sin(x)cos(pi/6)+cos(x)sin(pi/6)=1/2}}} Distribute



{{{sin(x)(sqrt(3)/2)+cos(x)(1/2)-sin(x)(sqrt(3)/2)+cos(x)(1/2)=1/2}}} Evaluate {{{cos(pi/6)}}} to get {{{sqrt(3)/2}}}. Evaluate {{{sin(pi/6)}}} to get {{{1/2}}}



{{{(sqrt(3)/2)sin(x)+(1/2)cos(x)-(sqrt(3)/2)sin(x)+(1/2)cos(x)=1/2}}} Rearrange the terms.



{{{cos(x)=1/2}}} Combine like terms.



{{{cos(x)=arccos(1/2)}}} Take the arccosine of both sides



{{{x=pi/3}}} or {{{x=-pi/3}}} Take the arccosine of 1/2 to get {{{pi/3}}} or {{{-pi/3}}}



Since the value {{{x=-pi/3}}} is NOT in the range [0,2pi), this means that {{{-pi/3}}} is NOT a solution



So the first solution is {{{x=pi/3}}}




Now add {{{2pi}}} to {{{pi/3}}} to get {{{2pi+pi/3=6pi/3+pi/3=(7pi)/3}}}. Take note how this value is outside the interval [0,2pi). So this value is NOT a solution



Now add {{{2pi}}} to {{{-pi/3}}} to get {{{2pi-pi/3=6pi/3-pi/3=(5pi)/3}}}. Since this value is inside the interval [0,2pi), this means that {{{(5pi)/3}}} is a solution




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Answer:



So the solutions are {{{x=pi/3}}} or {{{(5pi)/3}}}