If the equation  21cos−1x2π−21cos−1xπa+12−a2=0   has only one real solution then subsets of values of ‘a’ are

let  2πcos−1x =t   since  0≤cos−1x≤π⇒1π≤1cos−1x≤∞⇒1≤πcos−1x≤∞⇒2≤t<∞→122π/cos−1x−a+122π/cos−1x−a2=0⇒t2−a+12t−a2=0⇒t=a+12±a+122+4a22 From (1) 2≤a+12±a+122+4a22≤∞⇒4≤a+12±a+122+4a2≤∞⇒±a+122+4a2≥4−a+12=72−a⇒a+122+4a2≥72−a2⇒a2+14+a+4a2≥494+a2−7a⇒4a2+8a−12≥0⇒a2+2a−3≥0⇒(a−1)(a+3)≥0⇒a≤−3 or a≥1