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Okay, I got this problem wrong on my homework last night. I asked my sub (my teacher is on maternity leave), but he only got me more confused. If you could please show me your answer, as well as your process, I would appreciate the help!
5.5.4

The question is:
Find all the solutions of cos (x/2) - sin x = 0 in the interval [0,2pi]

Thank you for your time, John

A modified version of the double angle formulae has to be considered here:

sin (2x) = 2 sinx cos x is to be re-interpreted as

sin x   = 2 sin(x/2)  cos (x/2)

With this, cos(x/2) -sin x =0  becomes

cos(x/2) -  2 sin(x/2)  cos (x/2) =0

cos(x/2) [ 1- 2sin(x/2)] =0

This means cos (x/2) =0  ie  x/2 = pi/2 =====> x=pi

or  sin(x/2) =1/2   ie  x/2 = pi/6, 5 pi/6 =====> x= pi/3,  5pi/3   (shown)

Hope this clarifies. Peace.
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Calculus

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Frederick Koh

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