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 November 15th, 2016, 09:34 AM #1 Newbie   Joined: Oct 2015 From: huddersfield Posts: 16 Thanks: 1 Trigonometric identities Can anyone please help me with this question? Solve the following between the angles of 0 and 360 deg 4cot²x - 6 cosec x = -6 Last edited by skipjack; November 16th, 2016 at 06:35 AM.
 November 15th, 2016, 10:25 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,656 Thanks: 1327 note the identity ... $\cot^2{x} = \csc^2{x} - 1$ substitute $\csc^2{x} - 1$ for $\cot^2{x}$ and solve the resulting quadratic ...
 November 16th, 2016, 05:51 AM #3 Newbie   Joined: Oct 2015 From: huddersfield Posts: 16 Thanks: 1 Can you show me the first stages of substituting the identity? I understand how to solve the quadratic and also getting the angles between 0 and 360, but I do not understand how to substitute the identity. Last edited by skipjack; November 17th, 2016 at 05:56 AM.
 November 16th, 2016, 06:07 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,656 Thanks: 1327 $4\cot^2{x}-6\csc{x}=-6$ substitute ... $4(\csc^2{x}-1)-6\csc{x}=-6$ now, proceed ...
 November 24th, 2016, 12:21 PM #5 Newbie   Joined: Nov 2016 From: Canada Posts: 13 Thanks: 3 Math Focus: All Roots of the equation are cosec x = 1/2 & cosec x = 1 Therefore sin x = 2 & sin x =1 sin x = 2: no solution, because sin x belongs to [-1, 1] sin x = 1 => x = n*PI + (-1)power of n * PI/2 for all values on n belongs to Z
 November 24th, 2016, 09:17 PM #6 Global Moderator   Joined: Dec 2006 Posts: 18,154 Thanks: 1418 It's simpler to state that sin x = 1 implies x = $\pi$/2 + 2n$\pi$, where n is an integer.

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