To elaborate on the result that nadeem and bsg used,

 

We know from the fundamental theorem of algebra state and factor theorem that the polynomial

 

P(x) = a_nxⁿ+a_n-1x^n-1+...+a_o has exactly n roots in the complex plane, say x₁, x₂, ..., x_n

 

So, we can write P(x) = a_n (x-x₁) (x-x₂)...(x-x_n)

 

Now, if this expression goes to zero for x_n+1 which is not equal to any of the other roots, then the only possibility is that a_n = 0.

 

This means P(x) = 0 for all x.

sin^-1 (1-x) = /2 + 2sin^-1x

 

So sin(sin^-1 (1-x) ) = sin(/2 + 2sin^-1x) = cos(2sin^-1x)

 

Hence 1-x = 1-2sin²(2sin^-1x) = 1-2x²

 

the solutions are x = 0 and x=1/2

 

Now understand this point that x cannot be greater than zero.

 

The range of sin^-1 function is [-/2, /2]. If x>0, then sin^-1x>0 and

/2 + 2sin^-1x > /2

 

This is why x=1/2 is not an admissible solution.

 

good job nadeem. that's what i wanted to be brought out. thanx

 

rohit: i wudnt give a problem that needed computation, but one which would illustrate something useful. So I guarantee solving my qns will not be a waste of time. You may even stand to gain.