Find the number of real solutions of the equation below 
Find the number of real solutions of the equation 
  $$\sec(\theta) + \csc(\theta) = \sqrt{15}$$ 
  lying between $0$ and $\pi$.

My Approach :
Converted the equation in the form of $\sin\theta$ and $\cos\theta$ and eventually got $\sin\theta + \cos\theta = -\sqrt\frac{3}{5}$. Don't know how to proceed from here. Am I going wrong?
The answer given in my book is 4.
Can anyone help me?
 A: I got something other.
Let $\sin{\theta}+\cos{\theta}=t$.
Thus, $$\sin \theta \cos \theta=\frac{t^2-1}{2}$$ and we obtain
$$t=\frac{\sqrt{15}(t^2-1)}{2}$$ or
$$\sqrt{15}t^2-2t-\sqrt{15}=0,$$
which gives $t=\sqrt{\frac{5}{3}}$ or $t=-\frac{4}{\sqrt{15}}$ and since $\left|\sqrt{\frac{5}{3}}\right|<\sqrt2$ and $\left|-\frac{4}{\sqrt{15}}\right|<\sqrt2$, 
we obtain $2$ solutions.
Indeed, $$\sin\theta+\cos\theta=\sqrt{\frac{5}{3}}$$ gives
$$\sin\theta\cdot\frac{1}{\sqrt2}+\cos\theta\cdot\frac{1}{\sqrt2}=\sqrt{\frac{5}{6}}$$ or
$$\sin\left(\theta+45^{\circ}\right)=\sqrt{\frac{5}{6}}$$ and since $0^{\circ}<\theta<180^{\circ},$ we obtain
$$\theta_1=\arcsin\sqrt{\frac{5}{6}}-45^{\circ}$$ and $$\theta_2=-\arcsin\sqrt{\frac{5}{6}}+135^{\circ}.$$
The equation $$\sin\theta+\cos\theta=-\sqrt{\frac{16}{15}}$$ has no solutions for $0^{\circ}<\theta<180^{\circ}.$
A: I don't think that it is necessary to find the solutions explicitly. Actually this is a calculus problem.
Hint. In $D:=(0,\pi/2)\cup (\pi/2,\pi)$, the continuous and differentiable function 
$$f(x):=\sec(x) + \csc(x)=\frac{1}{\cos(x)} + \frac{1}{\sin(x)}=
\frac{\sin(x)+\cos(x)}{\sin(x)\cos(x)}.$$
Its derivative is
$$f'(x)=\frac{\sin^3(x)-\cos^3(x)}{\sin^2(x)\cos^2(x)}$$
which implies that its sign is positive if and only if $\sin(x)>\cos(x)$. 
Hence $f$ is strictly decreasing in $(0,\pi/4]$, and it is strictly increasing in $[\pi/4,\pi/2)$ and $(\pi/2,\pi)$.
Note also that $f(\pi/4)=2\sqrt{2}<\sqrt{15}$, and
$$\lim_{x\to 0^+}f(x)=\lim_{x\to \pi/2^-}f(x)=\lim_{x\to \pi^-}f(x)=+\infty,\quad \lim_{x\to \pi/2^+}f(x)=-\infty.$$
Hence, by the Intermediate Value Theorem, the equation $f(x)=\sqrt{15}$ has THREE solutions in $(0,\pi)$:
one in $(0,\pi/4)$, one in $(\pi/4,\pi/2)$, and one in $(\pi/2,\pi)$ (see WA).   
P.S. Since $f(x)=-f(3\pi/2-x)$, it is easy to verify that the number of solutions of $f(x)=\sqrt{15}$ in $(0,2\pi)$ is FOUR (see WA).
