Which of the following is not the value of y that satisfies the system of equations?

$co{s}^{-1}x+{\left(si{n}^{-1}y\right)}^2=\frac{p{\pi }^2}{4}$   ...(1)
$\left(co{s}^{-1}x\right){\left(si{n}^{-1}y\right)}^2=\frac{{\pi }^4}{16}$
Therefor equation with roots $co{s}^{-1}$ and ${\left(si{n}^{-1}y\right)}^2$ is
$x^2-\frac{p{\pi }^2}{4}x+\frac{{\pi }^4}{16}=0$ 
And for real values of x
$\frac{p^2{\pi }^4}{16}-\frac{4{\pi }^4}{16}\ge 0\Rightarrow p^2-4\ge 0$
$\Rightarrow p^2\ge 4\Rightarrow p\ge \pm 2$
For $y=1$ and $y=-1$
$$\begin{gathered} co{s}^{-1}x+{\left(si{n}^{-1}y\right)}^2=\frac{p{\pi }^2}{4}\Rightarrow co{s}^{-1}x+\frac{{\pi }^2}{4}=\frac{{\pi }^2}{2}\Rightarrow co{s}^{-1}x=\frac{{\pi }^2}{4} \\ \left(co{s}^{-1}x\right){\left(si{n}^{-1}y\right)}^2=\frac{{\pi }^4}{16}\Rightarrow co{s}^{-1}x\left(\frac{{\pi }^2}{2}\right)=\frac{{\pi }^4}{16}\Rightarrow co{s}^{-1}x=\frac{{\pi }^2}{4} \end{gathered}$$ 
Hence, $y=1$ and $y=-1$ are possible
And for $y=\frac{1}{2}$
$$\begin{gathered} co{s}^{-1}x+{\left(si{n}^{-1}y\right)}^2=\frac{p{\pi }^2}{4}\Rightarrow co{s}^{-1}x+\frac{{\pi }^2}{36}=\frac{{\pi }^2}{2}\Rightarrow co{s}^{-1}x=\frac{17{\pi }^2}{36} \\ \left(co{s}^{-1}x\right){\left(si{n}^{-1}y\right)}^2=\frac{{\pi }^4}{16}\Rightarrow \left(co{s}^{-1}x\right)\left(\frac{{\pi }^2}{36}\right)=\frac{{\pi }^4}{16}\Rightarrow co{s}^{-1}x=\frac{9{\pi }^2}{4} \end{gathered}$$
Hence, $y=\frac{1}{2}$ is not possible