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 .
