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  4. If 2 tan 2 x-5 sec x-1=0 has 7 different roots in [0, (n π/2)], n ∈ N, then greatest value of n is

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Q. If $2 \tan ^2 x-5 \sec x-1=0$ has 7 different roots in $\left[0, \frac{n \pi}{2}\right], n \in N$, then greatest value of $n$ is

Trigonometric Functions

Solution:

$2 \tan ^2 x-5 \sec x-1=0$
$\Rightarrow 2\left(\sec ^2 x-1\right)-5 \sec x-1=0$
$\Rightarrow 2 \sec ^2 x-5 \sec x-3=0$
$\Rightarrow \sec x=\frac{6}{2}, \frac{-1}{2}=3, \frac{-1}{2}$
$\Rightarrow \sec x=3 \left(\sec x \neq \frac{-1}{2}\right)$
image
$\Rightarrow \cos x=\frac{1}{3} \Rightarrow 7$ solutions in $\left[0, \frac{15 \pi}{2}\right]$
$\therefore n=15$