solve sin(2x)=0.5. between 0<x<2pi
1)Take the inverse sin to take x from the sin(2x):
2x=arcsin(0.5).
2)Evaluate arcsin(0.5) to get pi/6:
so 2x= pi/6
3)Dividing by 2 to simplify we get
x=pi/12.
4)To find the second solution we note that (pi/2)-(pi/12) =(5pi/12) is also a solution.
So x= (5pi/12)
5)Sin(2x) has a period of pi. So to find the rest of the solutions we add pi to our previous solutions.
So now x=pi/12, 5pi/12, 13pi/12 , 17pi/12
YZ
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