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Trigonometric equations

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Question

If $2 \tan^{2} x - 5 \sec x = 1$ for exactly 10 distinct values of $x \in [0, n π], n \in N$ then the greatest value of n is

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Solution

$2 (\sec^{2} x - 1) - 5 \sec x = 1 \Rightarrow 2 \sec^{2} x - 5 \sec x - 3 = 0$
$\Rightarrow (\sec x - 3) (2 \sec x + 1) = 0$
$\Rightarrow \sec x = 3 \Rightarrow \cos x = \frac{1}{3}$
Which gives two values of x in each of $[0, 2 π], (2 π, 4 π], (4 π, 6 π], (6 π, 8 π]$ and $t w o v a l u e s i n (8 π, 10 π]$
$∴$ greatest value of n = 10.

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