<s>[INST] 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?
 [/INST] 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 $0^{\circ}<\theta<180^{\circ}.$
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