Question

2tansqx-5secx-1 has 7 different roots in [0,npie by 2]find greatest value of n​

Answer 1

Given

$\sf{2tan {}^{2}x - 5sec \: x \: - 1 = 0 } \\ \\ \sf{since \: tan {}^{2}x = 1 - sec {}^{2}x} \\ \\ \rightarrow \sf{2(1 - sec {}^{2}x) - 5sec \: x \: - 1 = 0 } \\ \\ \rightarrow \: \sf{ - 2sec {}^{2}x + 2 - 5sec \: x \: - 1 = 0} \\ \\ \rightarrow \: \boxed{\sf{2sec {}^{2}x + 5sec \: x \: - 1 = 0 }}$

On comparing with ax²+bx+c = 0,

a = 2,b = 5 and c = 1

Discriminant,D :

$\sf{b {}^{2} - 4ac = (5) {}^{2} - 4(2)( - 1) } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf{25 + 8} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 33$

Now,

$\sf{sec \: x \: = \: \frac{ - b \pm \: \sqrt{d} }{2a} } \\ \\ \implies \: \sf{sec \: x \: = \frac{ - 5 \pm \: \sqrt{33} }{4} } \\ \\ \implies \: \sf{sec \: x \: = \frac{ \sqrt{33} - 5}{4} \: or \: \frac{ - (5 + \sqrt{33)} }{4} }$