Let `(b cos x)/(2 cos 2x-1)=(b + sin x)/((cos^(2) x-3 sin^(2) x) tan x), b in R`. Equation has solutions ifA. InfiniteB. depends upon the value of bC. twoD. none of these

Let `(b cos x)/(2 cos 2x-1)=(b + sin x)/((cos^(2) x-3 sin^(2) x) tan x

1.	Let `(b cos x)/(2 cos 2x-1)=(b + sin x)/((cos^(2) x-3 sin^(2) x) tan x), b in R`. Equation has solutions ifA. InfiniteB. depends upon the value of bC. twoD. none of these
Answer» Correct Answer - C `(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)`.