=a ‌⇒‌‌4(1−sin‌x)+sin‌x=asin‌x(1−sin‌x) ‌⇒‌‌4−4sin‌x+sin‌x=asin‌x−asin‌2x ‌⇒‌‌asin‌2x−(3+a)sin‌x+4=0 . . . (i) It is a quadratic equation in sin‌x, so ‌D≥0 ⇒‌(3+a)2−4×4a≥0 ⇒9+a2+6a−16a≥0 ⇒‌a2−10a+9≥0 ⇒‌(a−1)(a−9)≥0 ⇒‌a≥9‌‌ or ‌a≤−9 Now, at a =9, Eq. (i) becomes ‌9sin‌2x−12sin‌x+4‌=0 ⇒‌(3sin‌x−2)2‌=0 ⇒‌sin‌x‌=‌