12th Power

Let $\displaystyle \sin^{2}x=\alpha$ and $\displaystyle \cos^{2}x=\beta$ .
Now, we have $\displaystyle \alpha+\beta=1$ .
According to the question, $\displaystyle \alpha^{5}+\beta^{5}=\frac{11}{36}$ .
So, taking $\displaystyle \alpha=\frac{1}{2}+x$ and $\displaystyle \beta=\frac{1}{2}-x$ and after some simplifications, we get

$5x^{4}+\frac{5x^{2}}{2}+\frac{1}{16}=\frac{11}{36}$

which can be easily solved to yeild $\displaystyle x^{2}=\frac{1}{12}$ .
So,

$\left(\sin^{2}x\right)^{6}+\left(\cos^{2}x\right)^{6}=\alpha^{6}+\beta^{6}=\left(\frac{1}{2}+x\right)^{6}+\left(\frac{1}{2}-x\right)^{6}=2x^{6}+\frac{15x^{4}}{2}+\frac{15x^{2}}{8}+\frac{1}{32}=\frac{13}{54}$

Hence, making the answer $\displaystyle \boxed{67}$

Let $a=\sin^2 x$ and $b=\cos^2 x$ . Then

$\begin{aligned} a+b & = 1 \\ a^2+b^2 & = (a+b)^2 - 2ab = 1- 2{\color{#3D99F6}ab} = 1-2\color{#3D99F6} \alpha & \small \color{#3D99F6} \text{Let }\alpha = ab \\ a^3+b^3 & = (a+b)(a^2+b^2) - ab(a+b) = 1 - 3\alpha \\ a^4+b^4 & = 1 - 3\alpha - \alpha(1-2\alpha) = 1-4\alpha + 2\alpha^2 \\ a^5+b^5 & = 1-4\alpha + 2\alpha^2 - \alpha (1 - 3\alpha) = 1 - 5\alpha + 5\alpha^2 \end{aligned}$

Therefore,

$\begin{aligned} \sin^{10}x + \cos^{10}x & = 1 - 5\alpha + 5\alpha^2 = \frac {11}{36} \\ \implies 5\alpha^2 - 5\alpha + \frac {25}{36} & = 0 \\ \alpha^2 - \alpha + \frac 5{36} & = 0 \\ \left(\alpha - \frac 16\right) \left(\alpha - \frac 56\right) & = 0 \\ \implies \alpha & = \frac 16 & \small \color{#3D99F6} \text{Since }\alpha = \frac {\sin^2 2x}4 \le \frac 14 \end{aligned}$

And

$\begin{aligned} \sin^{12}x + \cos^{12}x & = a^6+b^6 \\ & = \frac {11}{36} - \alpha (1-4\alpha + 2\alpha^2) \\ & = \frac {11}{36} - \frac 16 \left(1-4\times \frac 16 + 2\times \frac 1{36} \right) \\ & = \frac {13}{54} \end{aligned}$

Hence $m+n = 13+54 = \boxed{67}$ .