"The number of solutions of the equation ${32^{{{\tan }^2}x}} + {32^{{{\sec }^2}x}} = 81,\,0 \le x \le {\pi \over 4}$ is :"

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Accepted Answer

"${(32)^{{{\tan }^2}x}} + {(32)^{{{\sec }^2}x}} = 81$

$\Rightarrow {(32)^{{{\tan }^2}x}} + {(32)^{1 + {{\tan }^2}x}} = 81$

$\Rightarrow {(32)^{{{\tan }^2}x}} = {{81} \over {33}}$

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."

The number of solutions of the equation ${32^{{{\tan }^2}x}} + {32^{{{\sec }^2}x}} = 81,\,0 \le x \le {\pi \over 4}$ is :

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The number of solutions of the equation ${32^{{{\tan }^2}x}} + {32^{{{\sec }^2}x}} = 81,\,0 \le x \le {\pi \over 4}$ is :

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