amparvardi wrote:

Solve the equation $\frac{1}{\sin x}+\frac{1}{\cos x}=\frac 1p$ where $p$ is a real parameter.
Discuss for which values of $p$ the equation has at least one real solution and determine the number of solutions in $[0, 2\pi)$ for a given $p.$

Let $\sin x+\cos x=t$ , we have $\sin x\cos x=\frac{t^2-1}{2}$ , thus we can rewrite the given equation as $t^2-1=2pt\ \cdots[*]$ .

Note that the range of $t=\sqrt{2}\sin \left(x+\frac{\pi}{4}\right)$ is $|t|\leq \sqrt{2}$ .

The number of solutions for $t$ , which is denoted by $N(t)$ , is equivalent to that of intersection points of the parabola $y=t^2-1\ (|t|\leq \sqrt{2}\ and\ t\neq 0)$ and the line $y=2pt$ ,

The value of the slope $2p$ is $\pm \frac{1}\sqrt{2}$ when the line passes through the point $(\pm \sqrt{2},\ 1)$ .

Remark that given a value of $t$ , the number of $x$ is when $|t|<\sqrt{2}$ , 2 solutions, when $|t|=\sqrt{2}$ , 1 solution,

therefore the number of solutions are:

$|p|<\frac{1}{2\sqrt{2}}\ \cdots N(t)=2\Longrightarrow N(x)=2+2=4$

$|p|=\frac{1}{2\sqrt{2}}\ \cdots N(t)=2\Longrightarrow N(x)=2+1=3$

$|p|>\frac{1}{2\sqrt{2}}\ \cdots N(t)=1\Longrightarrow N(x)=2$

Faster Solution:

$[*]\Longleftrightarrow t-\frac{1}{t}=2p\ \because t\neq 0$ , Consider the graphs of the curve $y=t-\frac{1}{t}\ (|t|\leq \sqrt{2}\ and\ t\neq 0)$ and the line $y=p$ . The graph of $y=t-\frac{1}{t}$ is like that $y=\log x,\ y=-\log (-x)$ , or odd function with the asymptote $y=t,\ t=0$ .