Question

The number of solutions of the equation,
Arcsinx=2arctanx (in principal values) is

Answer 1

My ans is 1 bt it"z not right...

Answer 2

@michele_laino

Answer 3

@astrophysics

Answer 4

@directrix @hero

Answer 5

is the answer 3 ?

Answer 6

x=n(pi) where n=0,1,2 since principal values.
i think u must have considered only zero..! right?

Answer 7

How do u know the value of arc tan π
N even x can be in between [-1,1]
The domain of sin inverse x is this only ..
U took x=pi which is never possible

Answer 8

@mayankdevnani

Answer 9

here is my reasoning:
we can write this:
\[\Large \sin \theta = x,\quad \tan \left( {\frac{\theta }{2}} \right) = x\]
furthermore:
\[\Large \cos \theta = \sqrt {1 - {x^2}} \]

Answer 10

Sir what I did was I converted both lhs and rhs in the form of arctanx

Answer 11

Like rhs would be arctan 2x/1-x^2
And lhs would be arctanx/√1-x^2

Answer 12

yes! correct, since by substitution, we get this equation:
\[\Large \tan \theta = \frac{{2\tan \left( {\theta /2} \right)}}{{1 - {{\left\{ {\tan \left( {\theta 2} \right)} \right\}}^2}}} = \frac{{2x}}{{1 - {x^2}}}\]

Answer 13

and therefore:
\[\Large \begin{gathered}
\frac{x}{{\sqrt {1 - {x^2}} }} = \tan \theta = \frac{{2\tan \left( {\theta /2} \right)}}{{1 - {{\left\{ {\tan \left( {\theta /2} \right)} \right\}}^2}}} = \frac{{2x}}{{1 - {x^2}}} \hfill \\
\hfill \\
\frac{x}{{\sqrt {1 - {x^2}} }} = \frac{{2x}}{{1 - {x^2}}} \hfill \\
\end{gathered} \]

Answer 14

Yep exactly....:)

Answer 15

When I solved this I got only 1 solution which is x=0

Answer 16

please wait, I'm checking such solution

Answer 17

we have these steps:
\[\Large \begin{gathered}
\frac{x}{{\sqrt {1 - {x^2}} }} = \frac{{2x}}{{1 - {x^2}}} \hfill \\
\hfill \\
\frac{x}{{\sqrt {1 - {x^2}} }} - \frac{{2x}}{{1 - {x^2}}} = 0 \hfill \\
\hfill \\
\frac{x}{{\sqrt {1 - {x^2}} }}\left( {1 - \frac{2}{{\sqrt {1 - {x^2}} }}} \right) = 0 \hfill \\
\end{gathered} \]

Answer 18

Yep right

Answer 19

Now 1-x^2=4 will not give us solution

Answer 20

Complex roots

Answer 21

the first equation, is:
\[\Large \frac{x}{{\sqrt {1 - {x^2}} }} = 0\]
whose solution, is \(x=0\), which is acceptable, since at x=0, the radical exists

Answer 22

Yep n we cannot find real roots from second equation

Answer 23

So and shud be 1 only...isn't it?

Answer 24

Ans*

Answer 25

that's right!
second equation is an impossible equation in the set of the real numbers

Answer 26

Sir BT ans key says 3

Answer 27

the equation \(\sin \theta = x\), has two principal solutions if \(x\) is fixed

Answer 28

Sir principal value for arc sin x
is [-π/2,π/2]