tan(x/2-π/2)=sqrt2 ?

[lait]

i got x=2arccot(-sqrt2)+2πn but it's wrong?

 

[Deleted member 4993]

i got x=2arccot(-sqrt2)+2πn - but it's wrong ?

How did you get that answer? Please share your work.

That answer can be further simplified - may that is what you need to supply!

 

[lait]

How did you get that answer? Please share your work.

That answer can be further simplified - may that is what you need to supply!

im not really sure how to explain it but I changed tan to -cot then divided the negative so it would be cot(x/2) = -sqrt2 and solved x like that

 

[topsquark]

im not really sure how to explain it but I changed tan to -cot then divided the negative so it would be cot(x/2) = -sqrt2 and solved x like that

You mostly have it. But I'd try this: (It's a bit more direct.):

\(\displaystyle tan \left ( \dfrac{x}{2} - \dfrac{\pi}{2} \right ) = \sqrt{2}\)

\(\displaystyle arctan \left [ tan \left ( \dfrac{x}{2} - \dfrac{\pi}{2} \right ) \right ] = arctan \left [ \sqrt{2} \right ] \)

\(\displaystyle \dfrac{x}{2} - \dfrac{\pi}{2} = arctan \left [ \sqrt{2} \right ] \)

Arctan will return an angle between \(\displaystyle - \dfrac{\pi}{2}\) and \(\displaystyle \dfrac{\pi}{2}\) so what does that give for \(\displaystyle arctan \left [ \sqrt{2} \right ] \) ?

-Dan

PS Notice that the tangent function has a period of \(\displaystyle \pi\) rad, not just \(\displaystyle 2 \pi\) rad.