Ranges in trigonometry How to find the range of the sum or difference of two trigonometric functions?
$2\sin x-3\cos x$

Before this whenever the question of range i have solved they were either single trigonometric function or if they were in pair then they were in form of squires (eg. $2\cos^2x + \sin^2x$) so i used to convert them into one function.
I don't know how to solve this type of problem, i think we here also we need to change the it into single function?

Let $\theta$ be an angle which satisfies $\cos \theta = \dfrac{2}{\sqrt{13}}$ and $\sin \theta = -\dfrac{3}{\sqrt{13}}$. Then we have:
$2\sin x - 3\cos x$
$= \sqrt{13}\left(\dfrac{2}{\sqrt{13}}\sin x - \dfrac{3}{\sqrt{13}}\cos x\right)$
$= \sqrt{13}\left(cos\theta\sin x + \sin\theta\cos x\right)$
$= \sqrt{13}\sin(x+\theta)$.
Do you know what the range of $\sqrt{13}\sin(x+\theta)$ is?
Using the same method, we can get the general result $A\sin x + B\cos x = \sqrt{A^2+B^2}\sin(x+\theta)$ where $\theta$ satisfies $\cos\theta = \dfrac{A}{\sqrt{A^2+B^2}}$ and $\sin\theta = \dfrac{B}{\sqrt{A^2+B^2}}$. See this pdf for details.
Thanks Freddy for suggesting to include that link in my answer.
