Question

What is the principal solution for tanx=sqrt{3}?

Answer 1

\[\LARGE x=\tan^{-1} 3\]
And range for tan inverse is -pi/2 to +pi/2
so it should be pi/3?

Answer 2

But the answer is pi/3 and 4pi/3
but 4pi/3 is not in the range for tan inverse function..why is it so?

Answer 3

I researched a bit,and found principal solution takes values from o to 2pi,but that is quite contradicting

Answer 4

I think principal means within the interval
\[\large [0 \ , 2\pi)\]

Answer 5

Well, think about it, both are solutions. And anything bigger than 2pi is coterminal to one of them.

Answer 6

yeah,but then x=tan inverse root 3 and tanx=sqrt{3} have different principal solutions?that is the main confusing question to me.

Answer 7

both seem same things

Answer 8

Well, for one thing... the solution was for this equation...
\[\large \tan \ x = \sqrt3\]and not
\[\large x = \tan^{-1}\sqrt{3}\]
It's jut like how this
\[\large x^2 = 4\]has two solutions but
\[\large x = \sqrt4\]
has only one.

Answer 9

but still both are the same things? :/

Answer 10

They're not. Because to much like the square root, the inverse tangent only becomes a function through a restricted domain.

Answer 11

what did this mean^^?

Answer 12

don't play with big words :|

Answer 13

It takes a deep understanding of functions :D
Take the squaring function...
\[\huge f:\mathbb{R}\rightarrow\mathbb{R}\]\[\huge f:x\rightarrow x^2\]

Answer 14

Domain is all real numbers right?

Answer 15

yes!

Answer 16

Now, the function becomes onto, if we restrict the range...
\[\huge f:\mathbb{R}\rightarrow [0,\infty)\]

Answer 17

We need this function to be onto (surjective) if we're to define an inverse relation.

Answer 18

I see! cleared a lot of things! thanks :)

Answer 19

@yrelhan4,this will clear your doubt too :)

Answer 20

I wasn't even finished yet :/ But lol, you seem to understand it now :D

Answer 21

yeah,:P