real root $\mbox{\boldmath $a_1$}$ of the polynomial $Q_5(t)$, and with arbitrary accuracy because $Q_5(t)$ is a strictly increasing polynomial function. This also determines $x^*=\sqrt{a_1}=1.570029 \ldots(> 1.57)$ as the first positive root of the polynomial $Q_{20}(x)$ defined at (\[Ineq\_29\]).

[**(A/II)**]{} [$ x\!\in\!\left(1.57,\nfrac{\pi}{2}\right)$]{} Let us define the function: $$\label{Ineq_35}
\begin{array}{rcl}
\hat{f}(x)\!\!&\!\!=\!\!&\!\! f \left(\nfrac{\pi}{2}\!-\!x\right)\!=\!1\!-\! 8 \left(\nfrac{\pi}{2}- x \right)^{\!2}\!+\!\hat{h}_1(x)\cos4 x\!+\!\hat{h}_2(x)\cos2 x\!+\!\hat{h}_3(x)\sin2 x\\[1.5 ex]
\!\!&\!\!=\!\!&\!\! 1-8 \left(\nfrac{\pi}{2}- x \right)^{\!2}-\cos4 x+8 \left(\nfrac{\pi}{2}- x \right)^{\!2} \cos2 x \\[1.5 ex]
\!\!&\!\! \!\!&\!\!+ \left(4 \left(\nfrac{\pi}{2}- x \right)-4 \left(\nfrac{8}{45} -a \left(\nfrac{\pi}{2}- x \right) \right) \right)\sin2 x,
\end{array}$$ where $x \!\in\!(0,c_1)$ for $c_1=\nfrac{\pi}{2}-1.57=\nfrac{\pi}{2}-\nfrac{157}{100}\left(=0.00079\ldots\right)$ and $\!\hat{h}_1(x)\!=\!-\!1\!<\!0, \hat{h}_2(x)=8\left(\nfrac{\pi}{2}- x \right)^{\!2}>0, \hat{h}_3(x)
=4\left(\nfrac{\pi}{2}- x \right)-4\left(\nfrac{8}{45}-a\left(\nfrac{\pi}{2}- x \right) \right)\!\left(\nfrac{\pi}{2}- x \right)^5$.

We are proving that the function $\hat{f}(x)>0$.

Again, it is important to find sign of the polynomial $\hat{h}_3(x)$. As we see, that polynomial is the polynomial of $7^{th}$ degree or $$\label{Ineq_36}
\begin{array}{rcl}
\hat{h}_3(x)=\hat{P}_7(x)\!\!&\!=\!&\!\!\!-\!\nfrac{32}{945}x^7\!+\!\nfrac{16}{135}\pi x^6\!+\!\left(\nfrac{32}{45}\!-\!\nfrac{8}{45}\pi^2\right)x^5\!+\!\left(-\nfrac{16}{9}\pi \!+\!\nfrac{4}{27}\pi^3\right)x^4 \\[1.5 ex]
\!\!&\! \!&\!\! \!+\!\left(\nfrac{16}{9}\pi^2\!-\!\nfrac{2}{27}\pi^4 \right)\!x^3\!+\!\left(-\nfrac{8}{9}\pi^3\!+\!\nfrac{1}{45}\pi^5 \right)\!x^2 \\[1.5 ex]
\!\!&\! \!&\!\!\!+\!\left(\!-\!4\!+\!\nfrac{2}{9}\pi^4\!-\!\nfrac{1}{270}\pi^6 \right)\!x+2\pi-\nfrac{1}{45} \pi^5+\nfrac{1}{3780}\pi^7.
\end{array}$$ Using the factorization of the polynomial $\hat{P}_7(x)$ we have $$\label{Ineq_37}
\begin{array}{rcl}
\hat{P}_7(x)\!\!&\!=\!&\!\! \nfrac{1}{3780}\left(-2 x+\pi \right)\Big(64 x^6-192\pi x^5+\left(240\pi^2-1344 \right) x^4                            \\[2.0 ex]
\!\!&\! \!&\!\! \!+\!\left(-\!160\pi^3\!+\!2688\pi\right)x^3\!+\!\left(60\pi^4\!-\!2016\pi^2\right)x^2\!+\!\left(-\!12\pi^5\!+\!672\pi^3\!\right)x \\[1.0 ex]
\!\!&\! \!&\!\! +\pi^6-84\pi^4+7560\Big) = \nfrac{1}{3780}\left(-2 x+\pi \right) \hat{P}_6(x),
\end{array}$$ where $$\label{Ineq_38}
\begin{array}{rcl}
\hat{P}_6(x)\!\!&\!=\!&\!\! 64 x^6-192\pi x^5+\left(240\pi^2-1344 \right) x^4+\left(-160\pi^3+2688\pi \right) 