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#1 2006-06-23 11:58:44

fusilli_jerry89
Member
Registered: 2006-06-23
Posts: 86

trig

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

√2 sin² -sin = 0    Solve for values of sin. 0<x<2pi

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#2 2006-06-23 12:52:58

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: trig

Is it sin(x) for each?

sin(x) (√2sin²x - 1) = 0

sin(x) = 0 or sin²x = √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 00:45:48

Kurre
Member
Registered: 2006-07-18
Posts: 280

Re: trig

fusilli_jerry89 wrote:

3pi/3.

3pi/3=Pi

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#4 2007-02-28 17:41:19

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: trig

Ricky wrote:

sin(x) (√2sin²x - 1) = 0

It’s actually

  wink

Thus the solutions are

Last edited by JaneFairfax (2007-02-28 17:52:30)


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