Question

The number of solutions of the equation $ta{n}^{2}x-se{c}^{10}x+1=0$ for $x\in \left(0,20\right)$ is equal to

Answer 1

Answer: 6

$se{c}^{2}x-se{c}^{10}x=0$
$se{c}^{2}x\left(1-se{c}^{8}x\right)=0$
$1-se{c}^{8}x=0$ as $se{c}^{2}x\ge 1$
$\Rightarrow se{c}^{8}x=1\Rightarrow co{s}^{8}x=1\Rightarrow cosx=\pm 1$
$\Rightarrow x=n\pi ,\forall n\in I$
$\Rightarrow x=\left\{\pi ,2\pi ,3\pi ,4\pi ,5\pi ,6\pi \right\}\in \left(0,20\right)$
So, the required number of roots $=6$