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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. Tags identities, trig Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathatesme Trigonometry 3 January 13th, 2014 04:20 AM taylor_1989_2012 Trigonometry 4 March 4th, 2013 02:05 AM swish Trigonometry 6 March 1st, 2012 05:18 PM Trigonometry 3 November 7th, 2011 12:41 AM Natalie89 Algebra 2 February 28th, 2011 09:00 AM

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