Predicting sign of a Trigonometric Function in Intervals. What are the different ways one can predict the sign of Trigonometric functions. The function may contain combination of different functions.
let us suppose I have a function f(x)= sin(2x) - cos(3x)
Now I want to predict the sign of this function in the interval [0,2π].
Is there any mechanical(doing some math calculations) approach or it is a matter of fact that you have to remember the graphs of both the functions and have a good idea of what will be the result (without actually calculating any thing). 
However I can predict the nature of the function f(x) using this graph [pink-->Resultant || green-->cos(3x) || violet-->sin(2x)] ]1
please admit if you have some shortcut or any trick to quickly predict the sign.
sorry for my bad English :( 
 A: You could use the product-sum identities.
In particular, let's try this identity:
$$
\cos(a) - \cos(b) = 
-2 \sin\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right).
$$
You have sine and cosine rather than two sines, but we can fix that
using the fact that 
$\sin(\theta) = \cos\left(\frac12\pi - \theta\right).$ Write
\begin{align}
\sin(2x) - \cos(3x) &= \cos\left(\tfrac12\pi - 2x\right) - \cos(3x) \\
&= -2 \sin\left(\frac{\frac12\pi - 2x + 3x}{2}\right)
      \sin\left(\frac{\frac12\pi - 2x - 3x}{2}\right) \\
&= 2 \sin\left(\tfrac12 x + \tfrac14\pi\right)
     \sin\left(\tfrac52 x - \tfrac14\pi\right).
\end{align}
So $\sin(2x) - \cos(3x)$ will change sign at each value of $x$
such that either $\sin\left(\tfrac12 x + \tfrac14\pi\right) = 0$
or $\sin\left(\tfrac52 x - \tfrac14\pi\right) = 0.$
Those equations should be relatively easy to solve.
Then evaluate $\sin(2x) - \cos(3x)$ at any other point
($x=0$ is an easy one in this case) to find the signs in each interval.
A: The solution that follows is the only way I can think of that solves your problem. It really doesn't look to be too useful though.
If $f(x)$ is continuous on $\mathbb R$, then a sign change will only occur about the points, $\xi$, where $f(\xi) = 0$.
If $f(x) = \sin 2x - \cos 3x$, then you need to solve 
\begin{align}
   f(x) &= 0\\
   \sin(2x) &= \cos(3x) \\
   2 \sin(x) \cos(x) &= \cos^3(x) - 3 \sin^2(x) \cos(x) \\
   2 \sin(x) &= \cos^2(x) - 3 \sin^2(x) \\
   2 \sin(x) &= 1 - 4 \sin^2(x) \\
   \sin x &\in
      \left\{
         -\dfrac 14 - \dfrac{\sqrt 5}{4},
         -\dfrac 14 + \dfrac{\sqrt 5}{4}
      \right\} \\
   \sin x &\in \{ \sin(-54^\circ),\; \sin(18^\circ) \}\\
   x &\in 360^{\circ}n +\{18^\circ, 162^\circ, 234^\circ, 306^\circ\}
\end{align}
Since the period of $f(x)$ is $360^\circ$, then 
f(x) is negative if 
$x \pmod{360^\circ} \in
   (0^\circ, 18^\circ) \cup
   (162^\circ, 234^\circ) \cup
   (306^\circ, 360^\circ)$
f(x) is positive if 
$x \pmod{360^\circ} \in
   (18^\circ, 162^\circ) \cup
   (234^\circ, 306^\circ)$
