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May 9th, 2017, 10:11 AM   #1
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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?
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May 9th, 2017, 10:14 AM   #2
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Originally Posted by Jimkeller1993 View Post
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?
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May 9th, 2017, 10:36 AM   #3
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No the equation I have is as it reads in the original post

(4cot^2) - (6cosec x) =-6
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May 9th, 2017, 11:32 AM   #4
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$$4\cot^2(x)-6\csc(x)=-6$$

$$4(\csc^2(x)-1)-6\csc(x)+6=0$$

$$4\csc^2(x)-6\csc(x)+2=0$$

$$(4\csc(x)-2)(\csc(x)-1)=0$$
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May 9th, 2017, 01:43 PM   #5
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Nice Greg, couldn't you also factor out the two and toss it? I believe that would be required of simplification.
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May 9th, 2017, 01:47 PM   #6
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Yes - toss the '2'. I often overlook such things.
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May 9th, 2017, 11:46 PM   #7
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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°.
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May 16th, 2017, 08:36 AM   #8
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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.
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May 17th, 2017, 06:35 AM   #9
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1 = sin(90°) = sin(90° + 360°), etc. For your problem, the multiple of 360° is zero.
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