Find the number of solutions of the trigonometric equation in $(0,\pi)$ Find the number of solutions of the equation $$\sec x+\csc x=\sqrt {15}$$ in $(0,\pi)$. The question is easy. But when you solve, you get would get $4$ as the answer. I am sure the method gives $4$ as answer, but the correct answer is $3$. I don't think there is an extraneous root of the equation. So, which of the $4$ solutions is not satisfying the equation? Thanks.
 A: Abbreviate $\sin x$ and $\cos x$ to $s$ and $c$.  Note that for $x\in(0,\pi)$, we have $s=\sqrt{1-c^2}$ (i.e., $\sin x\ge0$) and $-1\le c\le1$.  The equation is
$${1\over\sqrt{1-c^2}}+{1\over c}=\sqrt{15}$$
which implies
$${1\over1-c^2}=15-{2\sqrt{15}\over c}+{1\over c^2}$$
which can be rewritten as
$$c^2=(1-c^2)(15c^2-2\sqrt{15}c+1)$$
This is a quartic which indeed has four roots in the interval $-1\le c\le 1$.  However, let's go back and reiterate with emphasis:  the equation we want to solve implies $c$ is one of four values.  But the implication does not necessarily run in the other direction.  In particular, in order for a root of the quartic to be a solution of the original equation, we must also have
$${1\over\sqrt{1-c^2}}=\sqrt{15}-{1\over c}\ge0$$
which means we cannot have $0\lt c\lt1/\sqrt{15}$.  But writing the quartic as
$$P(c)=(c^2-1)(15c^2-2\sqrt{15}c+1)+c^2$$
we see that $P(0)=-1\lt0$ while $P(1/\sqrt{15})=1/15\gt0$, so one of the four roots (at least) does not solve the original equation.
A: $$\frac{1}{\sin x}+\frac{1}{\cos x}=\sqrt{15}\\\frac{\sin x +\cos x}{\sin x\cdot \cos x}=\sqrt{15}\\ $$use this substitution $\sin x +\cos x=u$ 
$$\sin x+\cos x =\sqrt{15}\sin x\cdot\cos x\\u^2=\sin^2 x +\cos^2x+2\sin x\cdot\cos x \\\frac{u^2-1}{2}=\sin x\cdot\cos x \\u=\sqrt{15}\frac{u^2-1}{2}$$ is this method that , you used to solve ?
A: If you set $X=\cos x$ and $Y=\sin X$, the equation is
$$
\frac{1}{Y}+\frac{1}{X}=\sqrt{15}
$$
to be considered together with $X^2+Y^2=1$.
Note that we must have $Y>0$, since $x\in(0,\pi)$.
The equation becomes $X+Y=\sqrt{15}\,XY$ and it's convenient to make another substitution, namely $S=X+Y$, $P=XY$, so we get
$$
\begin{cases}
S=P\sqrt{15}\\
S^2-2P=1
\end{cases}
$$
Thus we have $15P-2P-1=0$, that means
$$
P=\frac{1\pm4}{15}
$$
so $P=1/3$ or $P=-1/5$. If $P=1/3$, we have $S=\sqrt{15}/3$ and
$$
\begin{cases}
X+Y=\sqrt{15}/3\\
XY=1/3
\end{cases}
$$
This is solved by considering the roots of $3t^2-\sqrt{15}\,t+1=0$, giving
$$
X=\frac{\sqrt{15}-\sqrt{3}}{6},\qquad Y=\frac{\sqrt{15}+\sqrt{3}}{6}\\
\text{or}\\
X=\frac{\sqrt{15}+\sqrt{3}}{6},\qquad Y=\frac{\sqrt{15}-\sqrt{3}}{6}
$$
that correspond to two solutions in $(0,\pi)$.
For $P=-1/5$, we get $S=-\sqrt{15}/5$ and so
$$
\begin{cases}
X+Y=-\sqrt{15}/5\\
XY=-1/5
\end{cases}
$$
which is solved by considering $5t^2-\sqrt{15}\,t-1=0$, giving
$$
X=\frac{\sqrt{15}-\sqrt{20}}{10},\qquad Y=\frac{\sqrt{15}+\sqrt{20}}{10} \\
\text{or}\\
X=\frac{\sqrt{15}+\sqrt{20}}{10},\qquad Y=\frac{\sqrt{15}-\sqrt{20}}{10}
$$
and only the first one corresponds to an angle in $(0,\pi)$.
