$$ - 1 \le {{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi } \le 1$$

$$ \Rightarrow - {\pi \over 2} \le {\sin ^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right) \le {\pi \over 2}$$

$$ \Rightarrow - 1 \le {1 \over {4{x^2} - 1}} \le 1$$

$$\therefore$$ $${1 \over {4{x^2} - 1}} + 1 \ge 0$$

$$ \Rightarrow {{1 + 4{x^2} - 1} \over {4{x^2} - 1}} \ge 0$$

$$ \Rightarrow {{4{x^2}} \over {4{x^2} - 1}} \ge 0$$

$$ \Rightarrow $$ $${{4{x^2}} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0$$ ...... (1)

$$\therefore$$ $$x \in \left( { - \alpha , - {1 \over 2}} \right) \cup \{ 0\} \cup \left( {{1 \over 2},\alpha } \right)$$ .....(2)

And $${1 \over {4{x^2} - 1}} - 1 \le 0$$

$$ \Rightarrow {{1 - 4{x^2} + 1} \over {4{x^2} - 1}} \le 0$$

$$ \Rightarrow {{2 - 4{x^2}} \over {4{x^2} - 1}} \le 0$$

$$ \Rightarrow {{2{x^2} - 1} \over {4{x^2} - 1}} \ge 0$$

$$ \Rightarrow $$ $${{\left( {\sqrt 2 x + 1} \right)\left( {\sqrt 2 x - 1} \right)} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0$$ ...... (3)

$$x \in \left( { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left( { - {1 \over 2},{1 \over 2}} \right) \cup \left( {{1 \over {\sqrt 2 }},\alpha } \right)$$ .....(4)

From (3) and (4), we get

$$\therefore$$ $$x \in \left[ { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left[ {{1 \over {\sqrt 2 }},\alpha } \right) \cup \{ 0\} $$