Question 553053
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Since *[tex \Large \sin(2x)\ =\ 2\sin(x)cos(x)] we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin(x)cos(x)\ =\ cos(x)]


Multiply by *[tex \Large \frac{1}{\cos(x)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin(x)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \arcsin\left(\frac{1}{2}\right)]


Use the unit circle, recalling that sin of the angle is the *[tex \Large y]-coordinate of the point of intersection of the terminal ray with the unit circle and find all angles where *[tex \Large \sin(x)\ =\ \frac{1}{2}] in your given interval.  Note that the given interval is one and a half trips around the circle.  Hint: start at *[tex \Large \pi] and go backwards.


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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