For what values of 'b' does the equation `( b cos x)/( 2 cos 2x -1) =( b + sin x)/( (cos^(2)x -3 sin^(2) x) tan x )` possess solutions.

Options

A. Infinite
B. depends upon the value of b
C. two
D. none of these

Answer

C

Explanation

`(b cos x)/(2 cos 2x-1)=(b+sin x)/((cos^(2) x-3 sin^(2) x) tan x)`
`{:((1),2 cos 2x -1 ne 0,rArr,n ne npi pm pi/6", " n in Z),((2),tan x ne 0,rArr,x ne npi", " n in Z),((3),cos^(2) x-3 sin^(2) x ne 0,rArr,x ne npi pm pi/6", "n in Z):}`
Also `2 cos 2x-1 =2 (cos^(2) x-sin^(2) x)-(cos^(2) x+sin^(2)x)`
`= cos^(2) x-3 sin^(2) x`
Now, the given equation reduces to
`b sin x=b + sin x`
`rArr sin x=b/(b-1)`
Now, `-1 le sin x le 1`
`rArr -1 le b/(b-1) le 1`
`rArr b/(b-1) +1 ge 0` and `b/(b-1) -1 le 0`
`rArr (2b-1)/(b-1) ge 0` and `1/(b-1) le 0`
`rArr b in (-oo, 1/2] uu (1, oo)` and `b in (-oo, 1)`
`sin x ne 0, pm 1/2 or b ne 0, -1, 1/3`
When `b=1/2` then `sin x=-1`, which is not possible, as `tan x` is not defined for this value of `sin x`.
So, `b in (-oo, 1/2)-{-1, 0, 1/3}`
For any other value of b, `sin x` takes two values for `x in (0, pi)`.