f(x)=cos−1(π2sin−1(4x2−11))
Case I: −1≤4x2−11≤1
⇒4x2−1≥1 or 4x2−1≤−1 and 4x2−1=0
∴x2≥21 or x2≤0⇒x=0 and x=±21
∴x∈(−∞,−21]∪[21,∞)∪0
Case II: −1≤π2sin−1(4x2−11)≤1
⇒−2π≤sin−1(4x2−11)≤2π which is always true.
Hence, x∈(−∞,−21]∪[21,∞)∪0