IIT JEE , AIEEE Forum - Mathematics , Algebra

baba55

find no of solutions

sin invrse x = 2 tan invrse x

can graph b usd???

Decoder

infinite i think...

risin

sin-1x=sin-1(2x/1+x^2)=>x=0,-1,+1?

shikhar_destructo

@risin
only 0 is the soln.
@baba55
I will tell u the general approach.
First write the domain. There are a lot of qs wherein the solns are not acceptable as they do not lie in the domain or verify the solns at the end. First take sin or tan on both the sides
now expand LHS +or-x/(1-x2)^.5=(+or -) 2x/(1-x^2)
giving x=0 . If u accept 1 and -1 as soln substitute in the eqn
0=pi/4 absurd. so only 0 is the soln.

I hope it helps.

sriram.a

sin-1(x)=2tan-1(x)

tan-1(x/ $\sqrt{1-x^2}$ )=tan-1(x)

$\frac{x}{\sqrt{1-x^2}}=x$

$\Rightarrow$

$x=\sqrt{-3}$ .which is a absurd

if u find my answer correct/satisfactory plz vote me

akhil_o

$tan^{-1}x=A\\1/2sin^{-1}x=A\\sin^{-1}x=2A\\sin2A=x\\So\,the\,equation\,can \,be\,rewritten\,as\,$

$tanA=sin2A\\\frac{sinA}{cosA}=2sinAcosA\\sinA=0\,or\\cos^2A=1/2\\So\,we \,have\,$

$A=[n\pi,n\pi \pm\pi/4]\\x=tanA\\x=0\,or\,1\,or-1$

cross-checking values,

sin^-11=pi/2

tan^-11=pi/4

so sin^-11=2tan^-11

sin^-1-1=-pi/2

tan^-1-1=-pi/4

so sin^-1-1=2tan^-1-1

sin^-10=0

tan^-10=0

hence all 3 are solutions

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sin-1(x)=2tan-1(x)

tan-1(x/)=tan-1(x)

.which is a absurd

if u find my answer correct/satisfactory plz vote me

tan-1(x/ $\sqrt{1-x^2}$ )=tan-1(x)

$x=\sqrt{-3}$ .which is a absurd