May 9th, 2017, 10:11 AM  #1 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0  Trig identities
Solve the following equation for angles between 0 and 360 degrees 4cot^2  6 cosec x = 6 I think I need to make it a quadratic equation and find the angles from there by adding 90,180,270 and 360 to my two answers, how do I make this a quadratic term? Do I have to change the trig identities and if so what to? 
May 9th, 2017, 10:14 AM  #2  
Senior Member Joined: Aug 2012 Posts: 2,354 Thanks: 735  Quote:
 
May 9th, 2017, 10:36 AM  #3 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 
No the equation I have is as it reads in the original post (4cot^2)  (6cosec x) =6 
May 9th, 2017, 11:32 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,957 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
$$4\cot^2(x)6\csc(x)=6$$ $$4(\csc^2(x)1)6\csc(x)+6=0$$ $$4\csc^2(x)6\csc(x)+2=0$$ $$(4\csc(x)2)(\csc(x)1)=0$$ 
May 9th, 2017, 01:43 PM  #5 
Member Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 
Nice Greg, couldn't you also factor out the two and toss it? I believe that would be required of simplification.

May 9th, 2017, 01:47 PM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,957 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
Yes  toss the '2'. I often overlook such things. 
May 9th, 2017, 11:46 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,916 Thanks: 2199 
Multiplying 4cot²x  6cosec x = 6 by sin²x gives 4cos²x  6sin x = 6sin²x, so 4  4sin²x  6 sin x = 6sin²x, which implies sin²x  3sin x + 2 = 0. That can be factorized to give (sin x  2)(sin x  1) = 0. If x is real, sin x = 2 is impossible, so sin x = 1, which implies x = 90° + an integer multiple of 360°. 
May 16th, 2017, 08:36 AM  #8 
Newbie Joined: May 2017 From: alabama Posts: 6 Thanks: 0 
I don't quite follow that last bit, how did you work out the angle?
Last edited by skipjack; May 17th, 2017 at 06:30 AM. 
May 17th, 2017, 06:35 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,916 Thanks: 2199 
1 = sin(90°) = sin(90° + 360°), etc. For your problem, the multiple of 360° is zero.


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