Finding the least positive root How to find the least positive root of the equation $\cos 3x + \sin 5x = 0$?
My approach so far is to represent $\sin 5x$ as $\cos \biggl(\frac{\pi}{2} - 5x\biggr)$ then the whole equation reduces to $$2\cos \biggl(\frac{\pi}{4} - x\biggr)\cdot \cos \biggl(\frac{\pi}{4} - 4x\biggr) = 0$$
From here we can write:
$$\biggl(\frac{\pi}{4} - x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$
$$\biggl(\frac{\pi}{4} - 4x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$
Now there can be infinitely many solutions for this, what I am not getting how to compute the minimum among them? And what about if I am asked to find the maximum?
 A: You've already done the difficult part! Now just find all the solutions of $\cos(\pi/4-x)=0$ and $\cos(\pi/4-4x)=0$, and pick the smallest positive one.
A: EDIT: So apparently you want a full solution... 
A product of two real numbers is $0$ precisely when a factor is $0$, so you must have that 
$$ \begin{align}
& \cos\left(\tfrac{\pi}{4} - x\right) = 0 \qquad\;\; (1) \text{, or} \\\\
& \cos\left(\tfrac{\pi}{4} - 4x\right) = 0 \qquad (2).
\end{align} $$
Now we use that $\cos a = 0 \iff a \in \tfrac{\pi}{2} + \pi\mathbb{Z}$:
$$ \begin{align}
\cos \left(\tfrac{\pi}{4} - x\right) = 0 
  & \iff \tfrac{\pi}{4} - x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\
  & \iff x \in \tfrac{3\pi}{4} + \pi \mathbb{Z},
\end{align} $$
the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{4}$.
$$ \begin{align}
\cos \left(\tfrac{\pi}{4} - 4x\right) = 0 
  & \iff \tfrac{\pi}{4} - 4x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\
  & \iff 4x \in \tfrac{3\pi}{4} + \pi \mathbb{Z} \\\\
  & \iff x \in \tfrac{3\pi}{16} + \tfrac{\pi}{4}\mathbb{Z},
\end{align} $$
the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{16}$. The problem is now reduced to picking the smallest of two real numbers.
