Another trig identity: cosec^2 2x - cot 2x = 1

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JonnyK81
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Another trig identity: cosec^2 2x - cot 2x = 1

Stuck with the above. Needs to be solved in the range 0 < x < 180.

Just need to force it into a quadratic form... here are my attempts. Any ideas as to where I'm missing the plot?
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mqb2766
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On line 2, where does the double angle go when you subsitute for cosec^2(2x)
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JonnyK81
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(Original post by mqb2766)
On line 2, where does the double angle go when you subsitute for cosec^2(2x)
I've missed it.

Should be: 1 + cot^2 2x - cot 2x = 1
So: cot^2 2x - cot 2x = 0
So: cot 2x(cot 2x - 1) = 0
correct?
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mqb2766
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(Original post by JonnyK81)
I've missed it.

Should be: 1 + cot^2 2x - cot 2x = 1
So: cot^2 2x - cot 2x = 0
So: cot 2x(cot 2x - 1) = 0
correct?
Looks promising, just solve and sub back into the original equation to validate.
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JonnyK81
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(Original post by mqb2766)
Looks promising, just solve and sub back into the original equation to validate.
I know how to solve from here...

Use CAST diagrams.... just missed that 2 and it threw me right off.
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JonnyK81
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Thus cot 2x = 0 or cot 2x = 1

and 1/tan 2x = 0. or 1/tan 2x = 1

Thus, we can solve tan 2x = 1 in the range 0<x<180. The 1/tan 2x = 0 can be ignored as infinity is given when reciprocated?
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mqb2766
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(Original post by JonnyK81)
Thus cot 2x = 0 or cot 2x = 1

and 1/tan 2x = 0. or 1/tan 2x = 1

Thus, we can solve tan 2x = 1 in the range 0<x<180. The 1/tan 2x = 0 can be ignored as infinity is given when reciprocated?
cot(2x) = 0
gives valid solution(s).
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JonnyK81
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(Original post by mqb2766)
cot(2x) = 0
gives valid solution(s).
This thread springs to mind: https://www.thestudentroom.co.uk/sho....php?t=1901136
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mqb2766
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(Original post by JonnyK81)
This thread springs to mind: https://www.thestudentroom.co.uk/sho....php?t=1901136
Err yes. cos/sin = 0 so cos=0.
Not too hard?
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JonnyK81
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Solution for anyone who wants it:
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