Question

If the equation $\sec \theta +\text{cosec}\theta =c$ has two real roots between $0$ and $2\pi ,$ then the least integer which $c^2$ cannot exceed must be

Answer 1

Clearly, for $c=0,$ we get two solutions in $(0,2\pi )$ i.e., $x=\frac{3\pi }{4}$ and $x=\frac{7\pi }{4}$

$\sec \theta +\text{cosec}\theta =c,c\ne 0$
$\Rightarrow \sqrt{1+{\tan }^2\theta }+\sqrt{1+{\cot }^2\theta }=c$
$\Rightarrow \sqrt{1+{\lambda }^2}+\sqrt{1+\frac{1}{{\lambda }^2}}=c,$ where $\lambda =\tan \theta$
$\Rightarrow \sqrt{1+{\lambda }^2}\left(\frac{1+\lambda }{\lambda }\right)=c$
$\Rightarrow (1+\lambda {)}^2+2\lambda (1+{\lambda }^2)+{\lambda }^2=(c^2+1){\lambda }^2$
$\Rightarrow ({\lambda }^2+\lambda +1{)}^2=(c^2+1){\lambda }^2$
$\Rightarrow {\lambda }^2+\lambda +1+\lambda \sqrt{c^2+1}=0\cdots \left(1\right)$
or ${\lambda }^2+\lambda +1-\lambda \sqrt{c^2+1}=0\cdots \left(2\right)$

Discriminant of eqn. $\left(1\right)$ is
$D_1=(1+\sqrt{c^2+1}{)}^2-4$
$=(1+\sqrt{c^2+1}+2)(1+\sqrt{c^2+1}-2)$
$=(3+\sqrt{c^2+1})(\sqrt{c^2+1}-1)>0$

Discriminant of eqn. $\left(2\right)$ is
$D_2=(1-\sqrt{c^2+1}{)}^2-4$

$D_1>0\Rightarrow$ eqn. $\left(1\right)$ always has two real roots.
So, for $\sec \theta +\text{cosec}\theta =c$ to have only two real solutions,
$D_2<0$
$\Rightarrow (1-\sqrt{c^2+1}-2)(1-\sqrt{c^2+1}+2)<0$
$\Rightarrow (-\sqrt{c^2+1}-1)(3-\sqrt{c^2+1})<0$
$\Rightarrow 3-\sqrt{c^2+1}>0$
$\Rightarrow c^2+1<9$
$\Rightarrow c^2<8$

$\therefore 0\le c^2<8$