Find the range of $f(x)=2|{\sin x}|-3|\cos x|$ Find the range of $f(x)=2|{\sin x}|-3|\cos x|$
My attempt is as follows:-
$$f(x)=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|\right)$$
Let's assume $z=\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|$
$$f(x)=\sqrt{13}z$$
Let's find the range of $z^2$
$$z^2=\dfrac{4}{13}\cdot\sin^2x+\dfrac{9}{13}\cdot\cos^2x-\dfrac{6}
{13}|\sin2x|$$
$$z^2=1-\dfrac{6}{13}\cdot|\sin2x|$$
$$z^2=1-\dfrac{6}{13}\cdot[0,1]$$
$$z^2=1-\left[0,\dfrac{6}{13}\right]$$
$$z^2=\left[\dfrac{7}{13},1\right]$$
As $0$ is not there in the range of $z^2$, so we can say $z\in \left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]$
Hence $y \in \sqrt{13}\left(\left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]\right)$
$$y\in[-\sqrt{13},-\sqrt{7}]\cup[\sqrt{7},\sqrt{13}]$$
But actual answer is $[-3,2]$

Your domain is 0 to $\frac{\pi}{2}$. That's where sine and cosine are positive. So at those points you are limited to the range of $[-3,2]$. 
Unless you are given additional instructions, you are way over complicating it. 
