# 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)
D. [-1,1]
E. [2,infinity)
I tried everything I know.
I found only x=1
I need some ideas,suggestions..

#### 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)
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)
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.

#### 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! #### Attachments

• Naibu Shi

#### 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.

#### Naibu Shi

##### New member
The third line is wrong so the rest is not true.
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;+ ∞>

#### Naibu Shi

##### New member
The third line is wrong so the rest is not true.