Trigonometry
posted by bob .
sin (4x) = .5 what are the 4 solutions? How did you get them?

I saw your other post.
the way I do these is .....
treat 4x as your angle, ignore the fact it is a multiple of 4
by CAST rule
4x is in quadrants I and II since the sine is + in those quadrants
so 4x = 30° or 150°
(since sin30 = .5 and sin150 = .5)
if 4x = 30, then x = 30/4 or 7.5°
if 4x = 150, then x = 150/4 = 37.5°
now the period of sin 4x = 360°/4 = 90°
so by adding/subtracting any multiple of 90° we get other answers, that is
7.5+90= 97.5
37.5+90 = 127.5
97.5+90 = 187.5
127.5+90=217.5
187.5+90=277.5
217.5+90=307.5
adding another 90 will exceed the 360°
so this equation actually has 8 different answers
now recalling our previous question, I actually omitted 2 answers, there should have been 4.
I recall our angle was 2x, so the period would have been 360/2 = 180°
so by adding 180 to the 2 answers I gave you, will produce the other two remaining angles.
In general for both sin (kØ) and cos(kØ)
the period would be 360/k and there would be 2k solutions 
Okay that makes sense. I knew i was overlooking something. Thanks for the help Reiny! it really helped me out

You are welcome
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