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Trigonometric equations

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Question

If $2 \tan^{2} x - 5 \sec x = 1$ for exactly 7 distinct values of $x \in [0, \frac{n π}{2}], n \in N$ then the greatest value of n is

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Solution

We have    $2 \tan^{2} x - 5 \sec x = 1$
   $\begin{array}{l} \Rightarrow 2 (\sec^{2} x - 1) - 5 \sec x = 1 \\ \Rightarrow 2 \sec^{2} x - 5 \sec x - 3 = 0 \end{array}$
$\Rightarrow (\sec x - 3) (2 \sec x + 1) = 0$
$\Rightarrow \sec x = 3 o r s e c x = - \frac{1}{2}$
$\Rightarrow \cos x = \frac{1}{3}$
Which gives two values of x in each of   $[0, 2 π], (2 π, 4 π], (4 π, 6 π]$ and one value in $[6 π, 6 π + \frac{π}{2}]$ .
$∴$ greatest value of n = 15.

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