Official solution
\[{(32)^{{{\tan }^2}x}} + {(32)^{{{\sec }^2}x}} = 81\]
\[ \Rightarrow {(32)^{{{\tan }^2}x}} + {(32)^{1 + {{\tan }^2}x}} = 81\]
\[ \Rightarrow {(32)^{{{\tan }^2}x}} = {{81} \over {33}}\]
taking log of base 32 both side,
\[ \Rightarrow \] tan2x = \[{\log _{32}}\left( {{{81} \over {33}}} \right)\]
\[ \Rightarrow \] tan x = \[\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)} \]
As value of \[\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)} \] belongs to (0, 1).
In interval \[\,0 \le x \le {\pi \over 4}\] only one solution.