Expo-Trig equation

$\begin{aligned} 81^{\sin^2 x} + 81^{\cos^2 x} & = 30 \\ 81^{\sin^2 x} + 81^{1 - \sin^2 x} & = 30 \\ \color{#3D99F6}{81^{\sin^2 x}} + \frac {81}{\color{#3D99F6}{81^{\sin^2 x}}} & = 30 & \small \color{#3D99F6}{\text{Let } y = 81^{\sin^2 x}} \\ \color{#3D99F6}{y} + \frac {81}{\color{#3D99F6}{y}} & = 30 \\ \implies y^2 -30y + 81 & = 0 \\ (y-3)(y-27) & = 0 \end{aligned}$

$\implies \begin{cases} y = 3, & 81^{\sin^2 x} = 3^{4\sin^2 x} = 3^1, & 4\sin^2 x = 1, & \sin x = \frac 12, & x = \frac \pi 6 = A \\ y = 27, & 81^{\sin^2 x} = 3^{4\sin^2 x} = 3^3, & 4\sin^2 x = 3, & \sin x = \frac {\sqrt 3}2, & x = \frac \pi 3=B \end{cases}$

$\implies \dfrac {\pi}{A+B} = \dfrac {\pi}{\frac \pi 6 + \frac \pi 3} = \dfrac 1{\frac 16 + \frac 13} = \boxed{2}$

$\text {Multiply Both sides by } 81^{cos^2x} \\ \implies 81^{sin^2x+cos^2x}+81^{2cos^2x}=30\times81^{cos^2x}\\ = 81+81^{2cos^2x}=30\times81^{cos^2x}\\ \text {Put } 81^{cos^2x}=y \\ \implies 81+y^2=30y \\ =y^2-30y+81=0 \\= y^2-27y-3y+81=0\\ = (y-27)(y-3)=0 \\\implies 81^{cos^2x}=27,3 \\ = 3^{4cos^2x}=3^3,3^1\\ cos^2x=\dfrac34,\dfrac14\\ x=\dfrac{\pi}6,\dfrac{\pi}3$

$\large 81^{\sin^2{x}} + 81^{\cos^2{x}} = 30 \\ \text{Let } \cos^2x = a \\ \implies \sin^2 x = 1 - a \\ \therefore 81^{\sin^2{x}} + 81^{\cos^2{x}} = 30 \\ 81^{1-a} + 81^{a} = 30 \\ 81^1\times81^{-a} + 81^{a} = 30 \\ \dfrac{81}{81^a} + 81^a = 30 \\ \dfrac{81 + 81^{2a}}{81^a} = 30 \\ 81 + 81^{2a} = 30\times81^a \\ 81^{2a} - 30\times81^a+81 = 0 \\ \color{#3D99F6}{\text{Let }81^a = x} \\ \implies (81^x)^2 - 30\times81^a + 81 = 0 \\ x^2 - 30x + 81 = 0 \\ x^2 - 3x - 27x + 81 = 0 \\ x(x-3) - 27(x-3) = 0 \\ (x-27)(x-3) = 0 \\ \color{#D61F06}{x = 27, x = 3} \\ \implies 81^a = 27, 81^a = 3 \\ (3^{4})^a = 3^{3}, (3^4)^a = 3 \\ 3^{4a} = 3^3, 3^{4a} = 3 \\ \color{#3D99F6}{4a = 3, 4a = 1} \\ \color{#3D99F6}{\therefore \sin^2x = \dfrac{3}{4}, \sin^x = \dfrac{1}{4}} \\ \implies \sin x = \dfrac{\sqrt[]{3}}{4}, \sin x = \dfrac{1}{2} \\ \implies x = 60, x = 30\\ A = \dfrac{\pi}{3}; B = \dfrac{\pi}{6} \\ \color{#3D99F6}{\text{Therefore, the answer is } \dfrac{\pi}{\dfrac{\pi}{3} +\dfrac{\pi}{6}} = \dfrac{\pi}{0.5\pi} = \fbox{ 2 }}$

$81^{sin^2x} + 81^{cos^2x} = 30$

$81^{sin^2x} + 81^{1 - sin^2x} = 30$

$81^{sin^2x} + \frac {81}{ 81^{sin^2x} } = 30$

$\text {Let } y = 81^{sin^2x}$

Then:

$y + \frac {81}{y} = 30$

$y^2 - 30y + 81 = 0$

$(y - 3)(y - 27) = 0$

y = 3 or y = 27

Substituting these values back into

$y = 81^{sin^2x} = 3^{4sin^2x}$

we get:

$\sin {x} = ± \frac { \sqrt {3} }{2} \text { or } \sin {x} = ± \frac { 1 }{2}$

$\text {and solving it for } 0 ≤ x ≤ \frac {π}{2}$

$A = \frac {π}{3}$

$B = \frac {π}{6}$

Hence,

$\frac {π}{A + B} =\frac {π}{ \frac {π}{3} + \frac {π}{6} } = \frac {π}{ \frac {π}{2} } = \boxed {2}$