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 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
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 Originally Posted by Jimkeller1993 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?
Needs parens and clarity. Is that $4 \cot^2 (-6)$? Or is there an $x$ missing?

 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 $$4\cot^2(x)-6\csc(x)=-6$$ $$4(\csc^2(x)-1)-6\csc(x)+6=0$$ $$4\csc^2(x)-6\csc(x)+2=0$$ $$(4\csc(x)-2)(\csc(x)-1)=0$$ Thanks from phrack999
 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. Thanks from greg1313
 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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