Question

For $x\in [0,3\pi ]$, the number of integral values of $x$ satisfying ${\sec }^{2}x-4\tan x<0$ is

Answer 1

Answer: 3

${\sec }^{2}x-4\tan x<0$
$\Rightarrow 1+{\tan }^{2}x-4\tan x<0$

$\Rightarrow 2-\sqrt{3}<\tan x<2+\sqrt{3}$
$\Rightarrow n\pi +\frac{\pi }{12}<x<\frac{5\pi }{12}+n\pi ,n\in I$
$\Rightarrow \frac{\pi }{12}<x<\frac{5\pi }{12},\frac{13\pi }{12}<x<\frac{17\pi }{12},\frac{25\pi }{12}<x<\frac{29\pi }{12}$
$\therefore$ Integral values of $x$ can be $1,4$ and $7$ .