# find x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

#### Vali

##### Junior Member
I know x real number
I need to find "x" values such that:
2arctan(x) + arcsin(2x/(1+x^2))=pi
A. (0,infinity)
B. (-infinity,-1)U[1, infinity)
C. [1, infinity) right answer
D. [-1,1]
E. [2,infinity)
I tried everything I know.
I found only x=1
I need some ideas,suggestions..

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#### Steven G

##### Elite Member
I know x real number
I need to find "x" values such that:
2arctan(x) + arcsin(2x/(1+x^2))=pi
A. (0,infinity)
B. (-infinity,-1)U[1, infinity)
C. [1, infinity) right answer
D. [-1,1]
E. [2,infinity)
I tried everything I know.
I found only x=1
I need some ideas,suggestions..

I have not tried my ideas out (can't find any paper) but you can if you think they will be helpful

idea 1: Let u = 2arctan(x) and (pi - u) = arcsin(2x/(1+x^2))
idea 2: take derivative of both sides of
2arctan(x) + arcsin(2x/(1+x^2)) = pi

#### Dr.Peterson

##### Elite Member
I know x real number
I need to find "x" values such that:
2arctan(x) + arcsin(2x/(1+x^2))=pi
A. (0,infinity)
B. (-infinity,-1)U[1, infinity)
C. [1, infinity) right answer
D. [-1,1]
E. [2,infinity)
I tried everything I know.
I found only x=1
I need some ideas,suggestions..

I would start by defining u = arctan(x) and v = arcsin(2x/(1 + x^2)), so that 2u + v = pi. I would also find what tan(v) is, being very careful about signs and quadrants (since the given answers suggest that might be the entire point of the problem).

The double-angle formula for tangent will also be involved.

Give it a try, and show some work, so we can help you go further. At least, show what you already did, so we can find errors.

#### Vali

##### Junior Member
Thanks for the suggestions.
This is my try.If I would find the u interval for pi-2u then then the whole equation 2u+arcsin(sin(2u)) would be equal with pi for that unknown u interval.This way,I will find x: first value from u interval <= u (which is arctan(x) ) <= the last value from the interval.

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#### Dr.Peterson

##### Elite Member
Thanks for the suggestions.
This is my try.If I would find the u interval for pi-2u then then the whole equation 2u+arcsin(sin(2u)) would be equal with pi for that unknown u interval.This way,I will find x: first value from u interval <= u (which is arctan(x) ) <= the last value from the interval.

It's hard to follow some of what you wrote, but I think you wrote sin(2u) where you should have had tg(2u), and I'm not sure where you are going after that.

Here is what I did when I redid my work this time:

sin(v) = sin(pi - 2u) = sin(2u) = 2 sin(u) cos(u)

Fill in the left and right sides with what you know from the definitions of u and v, and solve for x. Then check domains and/or ranges.

#### Vali

##### Junior Member
I did it with the derivative.
Thank you for your suggestions!

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#### Naibu Shi

##### New member
I did it with the derivative.
Thank you for your suggestions!
im kinda curious about why x²-1>0
and yeah, its been a while since you did that so itd be totally fine if you didn't remember what u did

#### Steven G

##### Elite Member
im kinda curious about why x²-1>0
and yeah, its been a while since you did that so itd be totally fine if you didn't remember what u did
The third line is wrong so the rest is not true.
Please post your answer so we can verify if it is correct or not.

#### Naibu Shi

##### New member
The third line is wrong so the rest is not true.
Please post your answer so we can verify if it is correct or not.
Since you know that the derivate is 0 you find out that this is a constant function, this is the best i could do the thing is that when you graphic this, the function isnt a constant in its whole domain so i think this was close but i couldnt say why its constant only in when x ∈ <-∞;-1]U[1;+ ∞>

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#### Naibu Shi

##### New member
The third line is wrong so the rest is not true.
Please post your answer so we can verify if it is correct or not.
I DID IT
Well, this is not the answer you may be looking for cause i didnt prove that its π, i proved that the function is constant when x>1 but i hope it helps for your problem

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