p is not = 0. if $\frac{-1}{2\sqrt{2}} < p < \frac{1}{2\sqrt{2}}$ then there are exactly four solutions. one in second quadrant,one in fourth quadrant and the remaining two both in (either first or third ) quadrant. if $\ p=\frac{-1}{2\sqrt{2}} ,\frac{1}{2\sqrt{2}}$ exactly three solutions one in second quadrant,one in fourth quadrant and the remaining one in (either first or third ) quadrant. for all other p exactly two solutions. one in second quadrant,one in fourth quadrant

Square both sides of equation, simplify those, sketch both sides of equation to find out number of solutions where $y=p^{2}+x*0$ means contant function.

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 $y=t,\ t=0$.