Question

If $2{\tan }^{2}x-3\sec x=0$ has $5$ distinct solutions for $x\in [0,\frac{n\pi }{2}]$, n $\in$ N, then the least value of $n$ is

• A. $5$
• B. $7$
• C. $9$
• D. $11$

Answer 1

The correct option is C $9$

$2{\sec }^{2}x-3\sec x-2=0$
$\therefore \sec x=2$ i.e. $\cos x=\frac{1}{2}$
$\therefore$ there are atleast $5$ solution if $n\ge 9$
$\therefore$ least value of $n$ is $9$.