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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 1cos^2 x ? 8sin^3x6 sin x+1=0 
May 30th, 2013, 12:39 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716  Re: General Solution Quote:
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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