Let S be the set of all `alpha in R` such that the equation, `cos 2x + alpha sin x = 2alpha -7` has a solution. The S, is equal toA. RB. `[1,4]`C. `[3,7]`D. `[2,6]`

Let S be the set of all `alpha in R` such that the equation, `cos 2x +

1.	Let S be the set of all `alpha in R` such that the equation, `cos 2x + alpha sin x = 2alpha -7` has a solution. The S, is equal toA. RB. `[1,4]`C. `[3,7]`D. `[2,6]`
Answer» Correct Answer - D The given trigonometric equciton is `cos 2 x + alpha sin x = 2alpha -7` `rArr 1-2 sin^(2) x + alpha x = 2alpha - 7 " "[therefore 2x = 1- 2 sin^(2) x]` `rArr 2 sin^(2) - alpha sin x + 2alpha - 8 = 0 ` ` rArr 2 (sin ^(2) x - 4) - alpha (sin x - 2 ) = 0` ` rArr 2 (sin x-2) (sin x + 2) -alpha(sin x - 2)=0` `rArr (sin x-2) (2 sin x + 4-alpha) = 0 ` `therefore 2 sin x + 4 - alpha = 0" " [because sin x- 2 ne 0]` `rArr sin x = (alpha-4)/(2)" ".....(i)` Now, as we know `-1 le sin x le 1` `therefore " " -1 le (alpha-4)/(2) le 1" "["from Eq.(i)"]` `-2 le alpha - 4 le 2 rArr 2 le alpha le 6 rArr in [2,6]` and range of ` cos^(2) (3x) = [0,1]` So, the given equation holds if `1 + sin ^(4) x = 1 cos^(2)(3x)` `rArr sin^(4) x= 0 and cos^(2)3x =1` Since, `x in [-(5pi)/(2),(5pi)/(2)]` `therefore x = - 2pi, - pi,0, pi , 2pi` Thus, there are five different value of x is possible .