Solve the equation: $\;$ $3^{\log \left(\tan x\right)} - 2 \times 3^{\log \left(\cot x\right) + 1} = 1$
Given equation: $\;\;$ $3^{\log \left(\tan x\right)} - 2 \times 3^{\log \left(\cot x\right) + 1} = 1$
i.e. $\;$ $3^{\log \left(\tan x\right)} - 2 \times 3^{\log \left(\frac{1}{\tan x}\right)} \times 3^1 = 1$
i.e. $\;$ $3^{\log \left(\tan x\right)} - 6 \times 3^{\log \left(\tan x\right)^{-1}} = 1$
i.e. $\;$ $3^{\log \left(\tan x\right)} 6 \times 3^{- \log \left(\tan x\right)} = 1$
i.e. $\;$ $3^{\log \left(\tan x\right)} - \dfrac{6}{3^{\log \left(\tan x\right)}} = 1$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^{\log \left(\tan x\right)} = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p - \dfrac{6}{p} = 1$
i.e. $\;$ $p^2 - p - 6 = 0$
i.e. $\;$ $\left(p - 3\right) \left(p + 2\right) = 0$
i.e. $\;$ $p = 3$ $\;$ or $\;$ $p = -2$
Substituting the values of $p$ in equation $(2)$ give
when $\;$ $p = 3$,
$3^{\log \left(\tan x\right)} = 3 = 3^1$
$\implies$ $\log \left(\tan x\right) = 1$
$\implies$ $\tan x = 10^1 = 10$
i.e. $\;$ $x = \tan^{-1} \left(10\right) + n \pi, \;\; n \in Z$
when $\;$ $p = -2$,
$3^{\log \left(\tan x\right)} = -2$
$\implies$ $\log \left(\tan x\right) = \log_3 \left(-2\right)$
But logarithm of a negative number is not defined.
$\implies$ $p = -2$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\tan^{-1} \left(10\right) + n \pi, \;\; n \in Z \right\}$