Question

If `1/4` is a solution of `Sin2x=pCosx`, then p is? (a) -3 (b) 0 (c) 1 (d) -1 I think option (b) is correct, but I want solution.

Answer 1

`p=0.4948`

Explanation:

`sin2x=pcosx`

`hArr2sinxcosx=pcosx`

or `cosx(2sinx-p)=0`

i.e. either `cosx=0rArrx=(2n+1)pi/2`, where `n` is an integer - but this does not give us a value of `p`, as `1/4` is not among possible values.

or `2sinx-p=0` i.e. `p=2sinx`

as `1/4` is a solution, we have `2sin(1/4)-p=0` or `p=2sin(1/4)`

(assuming that `1/4` as solution is in radians)

`p=2xx0.2474=0.4948`