The roots of the equation `4x^2-2sqrt(5)x+1=0,a r e` (a)`sin36^0,sin18^0` (b) `sin18^0,cos36^0` (c)`sin36^0,cos18^0` (d) `cos18^0,cos36^0`

Explanation

`4x^2-2sqrt5x+1 = 0`

this is a quadratic equation, so its roots are,

`=> x= (-(-2sqrt5)+-sqrt((-2sqrt5)^2-4(4)(1)))/(2(4))`

`=>x = (2sqrt5+-sqrt4)/(8)`

`=>x= (sqrt5+-1)/4`

As `sin18^@ = (sqrt5-1)/4 and cos36^@ = (sqrt5+1)/4`

`=>x = sin18^@ and x = cos36^@`

So, option `(b)` is the correct option.