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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 =====>

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

Hope this clarifies. Peace.

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