SOLUTION: Find all solutions to the equation in the interval (0,2pi). sin(x+(pi/6))-sin(x-(pi/6))=1/2

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Question 773762: Find all solutions to the equation in the interval (0,2pi).
sin(x+(pi/6))-sin(x-(pi/6))=1/2
Found 2 solutions by KMST, tommyt3rd:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If we apply the trigonometric identities
sin%28a%2Bb%29+=+sin%28a%29cos%28b%29+%2B+cos%28a%29sin%28b%29
sin%28a-b%29+=+sin%28a%29cos%28b%29+-+cos%28a%29sin%28b%29
we get


With that, sin%28x%2Bpi%2F6%29-sin%28x-pi%2F6%29=1%2F2 turns into


%281%2F2%29%2Acos%28x%29%2B%281%2F2%29%2Acos%28x%29=1%2F2
cos%28x%29=1%2F2%29
We know that cos%28pi%2F3%29=1%2F2, so highlight%28x=pi%2F3%29 in quadrant I is a solution.
We know that there is an angle with the same cosine in quadrant IV.
I think of that angle as x=-pi%2F3, but the coterminal angle between 0 and 2pi is
x=2pi-pi%2F3 --> highlight%28x=5pi%2F3%29
That is the other solution.

Verification:
x=pi%2F3 --> system%28x%2Bpi%2F6=pi%2F3%2Bpi%2F6%2Cx-pi%2F6=pi%2F3-pi%2F6%29 --> system%28x%2Bpi%2F6=pi%2F2%2Cx-pi%2F6=pi%2F6%29 --> system%28sin%28x%2Bpi%2F6%29=1%2Csin%28x-pi%2F6%29=1%2F2%29 --> sin%28x%2Bpi%2F6%29-sin%28x-pi%2F6%29=1-1%2F2=1%2F2

x=5pi%2F3 --> system%28x%2Bpi%2F6=5pi%2F3%2Bpi%2F6%2Cx-pi%2F6=5pi%2F3-pi%2F6%29 --> system%28x%2Bpi%2F6=11pi%2F6%2Cx-pi%2F6=3pi%2F2%29 --> system%28sin%28x%2Bpi%2F6%29=-1%2F2%2Csin%28x-pi%2F6%29=-1%29 --> sin%28x%2Bpi%2F6%29-sin%28x-pi%2F6%29=-1%2F2-%28-1%29=-1%2F2%2B1=1%2F2

Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
sin(x+(pi/6))-sin(x-(pi/6))=1/2
sin(x)cos(pi/2)+cos(x)sin(pi/2)-[sin(x)cos(pi/2)-cos(x)sin(pi/2)]=1/2
2cos(x)*1/2=1/2
cos(x)=1/2

x=pi/3 or x=5*pi/3