The domain of the function cos^-1( 2 sin^-1( 1 4x^2-1) π ) is

$$f(x)={\mathrm{cos}}^{-1}(\frac{2{\mathrm{sin}}^{-1}(\frac{1}{4{x}^{2}-1})}{\pi })$$f(x)=cos−1(π2sin−1(4x2−11​)​)

Case I: $-1\leq \frac{1}{4{x}^{2}-1}\leq 1$−1≤4x2−11​≤1

$\Rightarrow 4{x}^{2}-1\geq 1$⇒4x2−1≥1 or $4{x}^{2}-1\leq -1$4x2−1≤−1 and $4{x}^{2}-1\neq 0$4x2−1=0

$\therefore {x}^{2}\geq \frac{1}{2}$∴x2≥21​ or ${x}^{2}\leq 0\Rightarrow x=0$x2≤0⇒x=0 and $x\neq \pm \frac{1}{2}$x=±21​

$$\therefore x\in (-\infty ,-\frac{1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},\infty )\cup {0}$$∴x∈(−∞,−2​1​]∪[2​1​,∞)∪0

Case II: $-1\leq \frac{2{\mathrm{sin}}^{-1}(\frac{1}{4{x}^{2}-1})}{\pi }\leq 1$−1≤π2sin−1(4x2−11​)​≤1

$\Rightarrow -\frac{\pi }{2}\leq {\mathrm{sin}}^{-1}(\frac{1}{4{x}^{2}-1})\leq \frac{\pi }{2}$⇒−2π​≤sin−1(4x2−11​)≤2π​ which is always true.

Hence, $x\in (-\infty ,-\frac{1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},\infty )\cup {0}$x∈(−∞,−2​1​]∪[2​1​,∞)∪0