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#1 2006-06-24 09:58:44

fusilli_jerry89
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trig

does anyone know how to get this question? I got pi/2 and 3pi/3.

√2 sin -sin = 0    Solve for values of sin. 0<x<2pi

 

#2 2006-06-24 10:52:58

Ricky
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Re: trig

Is it sin(x) for each?

sin(x) (√2sinx - 1) = 0

sin(x) = 0 or sinx = √2/2

x = 0, pi for the first, and sinx = √√2 / √2 which is 2^(3/4) / 2


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#3 2006-07-21 22:45:48

Kurre
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Re: trig

fusilli_jerry89 wrote:

3pi/3.

3pi/3=Pi

 

#4 2007-03-01 16:41:19

JaneFairfax
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Re: trig

Ricky wrote:

sin(x) (√2sinx - 1) = 0

Its actually

  wink

Thus the solutions are

Last edited by JaneFairfax (2007-03-01 16:52:30)


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