SOLUTION: Solve 4sin(π/6x)=2 for the four smallest positive solutions? to at least two decimal places I got .50,2.50,6.50,8.50 these are incorrect

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Question 1041035: Solve 4sin(π/6x)=2 for the four smallest positive solutions?
to at least two decimal places
I got .50,2.50,6.50,8.50
these are incorrect
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

4%2Asin%28expr%28pi%2F6%29%2Ax%29+=+2


%284%2Asin%28expr%28pi%2F6%29%2Ax%29%29%2F4+=+2%2F4 Divide both sides by 4


sin%28expr%28pi%2F6%29%2Ax%29+=+1%2F2 Simplify


expr%28pi%2F6%29%2Ax+=+pi%2F6+%2B+2pi%2An or expr%28pi%2F6%29%2Ax+=+5pi%2F6+%2B+2pi%2An Use the unit circle. Look for points on the circle that have a y coordinate of 1/2. Note: n is any integer.


expr%286%2Fpi%29%2Aexpr%28pi%2F6%29%2Ax+=+expr%286%2Fpi%29%2A%28pi%2F6+%2B+2pi%2An%29 or expr%286%2Fpi%29%2Aexpr%28pi%2F6%29%2Ax+=+expr%286%2Fpi%29%2A%285pi%2F6+%2B+2pi%2An%29 Multiply both sides by 6%2Fpi


x+=+1%2B12n or x+=+5%2B12n Simplify


The complete solution for x is x+=+1%2B12n or x+=+5%2B12n where n is any integer. This will generate an infinite number of solutions.


The first four positive solutions are when n = 0 and n = 1


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When n = 0, then


x+=+1%2B12n or x+=+5%2B12n


x+=+1%2B12%2A0 or x+=+5%2B12%2A0


x+=+1 or x+=+5


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When n = 1, then


x+=+1%2B12n or x+=+5%2B12n


x+=+1%2B12%2A1 or x+=+5%2B12%2A1


x+=+13 or x+=+17


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The four smallest positive solutions are therefore: 1, 5, 13, 17