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May 30th, 2013, 10:48 AM   #1
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General Solution

Can someone please help me solve this general solution, do I change 1 into 1-cos^2 x ?

8sin^3x-6 sin x+1=0
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May 30th, 2013, 12:39 PM   #2
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Re: General Solution

Quote:
Originally Posted by AzraaBux
Can someone please help me solve this general solution, do I change 1 into 1-cos^2 x ?

8sin^3x-6 sin x+1=0
Why would you?

Let u = sinx and you have a cubic in u.
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May 30th, 2013, 01:13 PM   #3
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Re: General Solution

I'm sorry but I don't understand that ?
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May 30th, 2013, 01:37 PM   #4
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Re: General Solution

The sin^3x does not seem to be operating anything.
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May 31st, 2013, 07:13 AM   #5
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8sin(x) - 6sin(x) + 1 = 0 implies 1/2 = 3sin(x) - 4sin(x) = sin(3x).
Hence ?/2 ?/3 + 2k? = 3x, where k is an integer, and so x = ?/6 ?/9 + 2k?/3.
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