Given, cos2x+7=1(2−sinx) 1−2sin2x+7=2a−asinx ⇒2sin2x−asinx+(2a−8)=0 This is a quadratic equation in sinx. Using the quadratic formula, sinx=[a±√{(−a)2−8(2a−3)}]∕(2×2) =[a±(a−8)]∕4 sinx=[a+(a−8)]∕4,sinx=[a−(a−8)]∕4 sinx=(2a−8)∕4,sinx=8∕4 sinx=(a−4)∕2,sinx=2, is not possible Since we know that −1≤sinx≤1 Therefore, −1≤(a−4)∕2≤1 Multiplying by 2 , −2≤a−4≤2 Adding 4 2≤a≤6