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Letting $t=\sin^2x$ , we are asked to minimise $\frac{t^3+(1-t)^3+1}{t^2+(1-t)^2+1}=\frac{3t^2-3t+2}{2t^2-2t+2}=\frac{3}{2}-\frac{1}{2(t^2-t+1)}$ for $0\leq t\leq 1$ . The minimum $a=\frac{5}{6}$ is attained at $t=\frac{1}{2}$ , when $t^2-t$ is minimal. Thus the answer is $\frac{1}{a}=\boxed{1.2}$ .

Consider

$\begin{aligned} \color{#3D99F6} (\sin^2 x + \cos^2 x)^3 & = \sin^6 x + 3\sin^4 x \cos^2 x + 3\sin^2 x \cos^4 x + \cos^6 x & \small \color{#3D99F6} \text{Recall }\sin^2 x + \cos^2 x = 1 \\ \color{#3D99F6} 1^3 & = \sin^6 x + 3\sin^2 x \cos^2 x {\color{#3D99F6}(\sin^2 x + \cos^2 x)} + \cos^6 x \\ & = \sin^6 x + 3\sin^2 x \cos^2 x + \cos^6 x \\ \implies \sin^6 x + \cos^6 x & = 1 - 3\sin^2 x \cos^2 x & \small \color{#3D99F6} \text{Similarly } \\ \implies \sin^4 x + \cos^4 x & = 1 - 2\sin^2 x \cos^2 x \end{aligned}$

Hence we have:

$\begin{aligned} X & = \frac {\sin^6 x+\cos^6 x+1}{\sin^4 x+\cos^4 x+1} \\ & = \frac {2-3\sin^2 x\cos^2 x}{2-2\sin^2 x\cos^2 x} \\ & = 1 - \frac {\sin^2 x\cos^2 x}{2-2\sin^2 x\cos^2 x} \\ & = 1 - \frac 1{{\color{#3D99F6}\frac 2{\sin^2 x\cos^2 x}}-2} & \small \color{#3D99F6} \text{Since }\sin 2 \theta = 2\sin \theta \cos \theta \\ & = 1 - \frac 1{{\color{#3D99F6}\frac 8{\sin^2 2 x}}-2} \end{aligned}$

Note that $X$ is minimum, when $\left|\dfrac 8{\sin^2 2x} - 2\right|$ is minimum. That is when $\sin^2 2x = 1$ and $\left|\dfrac 8{\sin^2 2x} - 2\right|=6$ , $\implies \min(X) = 1 - \dfrac 16 = \dfrac 56$ . Therefore, $a=\dfrac 56$ $\implies \dfrac 1a = \dfrac 65 = \boxed{1.2}$ .

$\sin^6x+\cos^6x+1=(\sin^2x+\cos^2x)^3-3(\cos^4x\sin^2x+\cos^2x\sin^4x)+1=2-3(\sin^2x+\cos^2x)\sin^2x\cos^2x=2-3(\sin x\cos x)^2$

$\sin^4x+\cos^4x+1=(\sin^2x+\cos^2x)^2-2\sin^2 x\cos^2 x+1=2-2(\sin x\cos x)^2$

Let $(\sin x\cos x)^2=y$ , then the fraction we have to minimize is equal to $\dfrac{2-3y}{2-2y}$

$y=(\sin x\cos x)^2=\dfrac{\sin^2(2x)}4$ , so $0\le y\le \frac14$

Since $\dfrac{2-3y}{2-2y}$ decreases when $0\le y\le \frac14$ , we just have to put $y=\frac14$ to get the minimum which is equal to $\frac56$

$\frac{\sin ^6(x)+\cos ^6(x)+1}{\sin ^4(x)+\cos ^4(x)+1} \Longrightarrow \frac{3 \cos (4 x)+13}{2 (\cos (4 x)+7)}$

Realizing at I could substitute a variable for $\cos 4x$ and restrict that variable's values to the interval [-1,1], then I realized that the expression was increasing through that interval and the minimum would be at the beginning of the interval, I evaluated the expression at that value and received the value of $\frac56$ . Therefore, the desired value is $\frac65$ or 1.2.

Using “keisan Advance Calculator “ in degrees.

x=10, 10, 8, 40<x<50, for minimum.

x= 40,1,50. x=.8333333=n.

1/n=1.2