Question

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

Answer 1

$sec^{2}x-sec^{10}x=0$
$sec^{2}x\left(1 - s e c^{8} x\right)=0$
$1-sec^{8}x=0$ as $sec^{2}x\geq 1$
$\Rightarrow sec^{8}x=1\Rightarrow cos^{8}x=1\Rightarrow cosx=\pm1$
$\Rightarrow x=n\pi ,\forall n\in Ι$
$\Rightarrow x=\left\{\pi , 2 \pi , 3 \pi , 4 \pi , 5 \pi , 6 \pi \right\}\in \left(0,20\right)$
So, the required number of roots $=6$