Question

The number of solution(s) of the equation ${\tan }^{2}x-{\sec }^{10}x+1=0$ in $x\in [0,60]$ is

Answer 1

${\tan }^{2}x-{\sec }^{10}x+1=0$
$\Rightarrow {\sec }^{2}x-{\sec }^{10}x=0$
$\Rightarrow {\sec }^{2}x(1-{\sec }^{8}x)=0$
$\Rightarrow {\sec }^{8}x=1(\because {\sec }^{2}x\ne 0)$
$\Rightarrow {\cos }^{8}x=1\Rightarrow \cos x=\pm 1\Rightarrow x=n\pi ,n\in Z$
Since $x\in [0,60]$
$\therefore x=0,\pi ,2\pi ,...,19\pi$

Hence, the required number of solutions are $20$.