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  #1  
Old November 18th, 2008, 02:17 PM
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Default Trigonometric function
Hello guys!

I have a problem solving this equation. I really dont know how to start Could someone pls help me with this...

The problem is that Find all solutions of the given equation in the indicated interval.

3sin2X+cosX = 0 [-pi/2, pi/2]

Thanks in advanced.
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  #2  
Old November 18th, 2008, 02:32 PM
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Hi,

Hint : \sin (2x)=2\sin x\cos x.
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Old November 18th, 2008, 02:56 PM
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thank you very much for replying.

OK so using that identity my equation

from 3sin2X+cox

will become (3)(2)sinxcosx + cosx = 0

6sinxcosx + cox = 0

Im I doing it right?
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Old November 18th, 2008, 03:01 PM
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Quote:
Originally Posted by cjru View Post
thank you very much for replying.

OK so using that identity my equation

from 3sin2X+cox

will become (3)(2)sinxcosx + cosx = 0

6sinxcosx + cox = 0

Im I doing it right?
Yes then factor out a cos x
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Old November 18th, 2008, 03:13 PM
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Originally Posted by 11rdc11 View Post
Yes then factor out a cos x
ok then I get cosX (6sinX) = 0

Therefore Cos X = 0 where in Cos can only be zero at x equal to pi/2 and -pi/2

Is this right?

How about the 6sinx = 0

Pls check if i did it right....

sin x = 0/6 then sin x = 0

Therefore sin x = 0 wherein sin can only be zero at x equal to 0 and pi.
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Old November 18th, 2008, 03:39 PM
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not quite ...

\cos{x}(6\sin{x} + 1) = 0
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Old November 18th, 2008, 06:53 PM
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o yeah! I over look it. So what will be the solution for 6sinX+1?

If I equate that to zero sinx = 1/6 Is this correct?
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Old November 18th, 2008, 09:20 PM
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o yeah! I over look it. So what will be the solution for 6sinX+1?

If I equate that to zero sinx = 1/6 Is this correct?
\sin{x} = -~\frac{1}{6}

x = \sin^{-1} {\bigg(-~\frac{1}{6}\bigg)}
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Old November 18th, 2008, 10:32 PM
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Thank you so much. Math help forum is the best.

God bless and more power to you guys!

Again thank you
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