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 May 30th, 2013, 10:48 AM #1 Newbie   Joined: Mar 2013 Posts: 14 Thanks: 0 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
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.

 May 30th, 2013, 01:13 PM #3 Newbie   Joined: Mar 2013 Posts: 14 Thanks: 0 Re: General Solution I'm sorry but I don't understand that ?
 May 30th, 2013, 01:37 PM #4 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: General Solution The sin^3x does not seem to be operating anything.
 May 31st, 2013, 07:13 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 8sin³(x) - 6sin(x) + 1 = 0 implies 1/2 = 3sin(x) - 4sin³(x) = sin(3x). Hence ?/2 ± ?/3 + 2k? = 3x, where k is an integer, and so x = ?/6 ± ?/9 + 2k?/3.

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