What is the smallest positive x satisfying _ x x+ _ x x=2 ?

Let $t = \log_{\sin x}\cos x$.

The given equation becomes $t + \dfrac{1}{t} = 2$.

$t^2 - 2t + 1 = 0$

$(t - 1)^2 = 0 \Rightarrow t = 1$

$\log_{\sin x}\cos x = 1$

$\cos x = \sin x$

$\tan x = 1$

The smallest positive value of $x$ is $\dfrac{\pi}{4}$.

For $x = \dfrac{\pi}{4}$, $\sin x = \dfrac{1}{\sqrt{2}}$ and $\cos x = \dfrac{1}{\sqrt{2}}$, which satisfy the conditions for the base and argument of logarithms (positive and not equal to $1$).

Answer: $\dfrac{\pi}{4}$