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

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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Reply 1

mqb2766

On line 2, where does the double angle go when you subsitute for cosec^2(2x)

Reply 2

JonnyK81

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?

Reply 3

mqb2766

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.

Reply 4

JonnyK81

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.

Reply 5

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?

Reply 6

mqb2766

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?