Let us rewrite the above equation as:

$\Large 81^{\sin^{2}x} + 81^{1-sin^{2}x} = 81^{\sin^{2}x} + \frac{81}{81^{sin^{2}x}} = 30$ .

Let $u = 81^{sin^{2}x}$ so that we obtain the quadratic equation $u + \frac{81}{u} = 30 \Rightarrow u^2 - 30u + 81=0 \Rightarrow (u-3)(u-27) = 0 \Rightarrow u = 3, 27$ . Now if $81^{sin^{2}x} = 3^{4\sin^{2}x} = 3, 27 \Rightarrow \sin^{2}x = \frac{1}{4}, \frac{3}{4} \Rightarrow \sin(x) = \frac{1}{2}, \frac{\sqrt{3}}{2}.$ Since $\frac{1}{2} < \frac{\sqrt{3}}{2},$ the answer is $\boxed{\frac{1}{2}}.$

Step 1 : Stare at the equation for 30 seconds.

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

Step 2 : notice that both 30 and 81(i.e. $3^{4}$ ) are related to each other (factors of three).

Step 3 : re-write the equation as:

$\boxed{3^{4sin^{2}x}+3^{4cos^{2}x}=30}$

Step 4 : let ☆ be $4sin^{2}x$ and ✿ be $4cos^{2}x$

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And re-write the equation as:

$\boxed{3^{☆}+3^{✿}=30}$

Step 4.99 : just by looking at it we can conclude that either ☆=3 ,✿=1 or vice versa

as $\boxed{3^3+3^1=30}$ = $27+3$ =30

Or

$\boxed{3^1+3^3=30}$ =3+27=30

Step 5 : bringing back sin and cos

If $4sin^{2}x=3$ ,then principal value of sin x will be sin 60° (i.e. $\dfrac{\sqrt{3}}{2}$ ) and cosx will cos 30°(I.e. $\dfrac{1}{2}$ )

If $4sin^{2}x=1$ then principal value of sinx will be sin 30°(I.e. $\dfrac{1}{2}$ ) and cos x will be cos 60°(I.e. $\dfrac{\sqrt{3}}{2}$ ) Hence the minimum value of sinx =sin30°(I.e. $\dfrac{1}{2}$ ) =0.5

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Note that we are neglecting the negative values of sinx and cosx as we are required to find the minimum value and negative values of sinx lies in third and fourth quadrants which will definitely be greater than 30°.

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Hence $\textcolor{#3D99F6}{0.5}$ is the answer!

Boom!