Find domain and range of $\log_2\left(\sin(2x)+\cos(2x)\right)$ I'm having trouble finding domain and range of this function.
Can somebody give me a HINT please? Thanks.
 A: $$
\log_2(\sin(2x)+\cos(2x)),
\\
\bbox[lightgreen]
{
(\sin(2x)+\cos(2x))>0,
\\
-1\le\sin(2x)\le{1},
\\
-1\le\cos(2x)\le{1}
}.
$$
Also don't forget to use this formula:
$$
\bbox[pink]
{
a\sin(x)+b\cos(x)= \sqrt{a^2+b^2}\sin\left(x+\arctan\left(\frac{b}{a}\right)\right)
}.
$$
A: \begin{align}
\cos(2x)+\sin(2x)
&=\cos(2x)\,r\cos \phi + \sin(2x)\,r\sin \phi \\
&= r\cos(2x-\phi)
\end{align}
\begin{cases}
r\cos \phi =1\\
r\sin \phi =1
\end{cases}
Finding $r$:
\begin{align}
r^2 (\sin^2\phi +\cos^2 \phi) &= 1^2+1^2\\
r &=\sqrt{2}
\end{align}
Finding $\phi$:
\begin{align}
\frac{r\sin \phi}{r\cos \phi} &= \frac{1}{1}\\
\tan \phi &= 1\\
\phi &= 45^\circ
\end{align}
Return to the main. Argument of logarithm must be positive.
\begin{align}
\cos(2x)+\sin(2x)&>0\\
\sqrt{2}\cos(2x-45^\circ) &>0\\
\cos(2x-45^\circ) &>0
\end{align}
\begin{align}
-90^\circ + 360^\circ k < 2x-45^\circ < 90^\circ + 360^\circ k\\
-45^\circ + 360^\circ k < 2x < 135^\circ + 360^\circ k\\
-22.5^\circ + 180^\circ k < x < 67.5^\circ + 180^\circ k
\end{align}
A: Hint: one way to find the range can be to try expressing $sin(2x)+cos(2x)$ as a single (transformed) sine function
