Solving $\cos3x=-\sin3x$, for $x \in [0,2\pi]$ For my first step, I would choose to either divide both sides or add sin3x to both sides. My professor in Pre-Calc told me I shouldn't divide any trig functions from either side. Despite this, I can only find a solution this way.
$$\frac{\cos3x}{-\sin3x}=1$$
$$-\tan3x=1$$
$$\tan3x=-1$$
Which would lead me to my method of solving trig equations which is:
$$-\tan\,\epsilon\,Q2\quad-\tan\,\epsilon\,Q4$$
$$\tan x=1 \, at\,\frac{\pi}{4}$$
$$Q2:\pi-\frac{\pi}{4}=\frac{3\pi}{4}$$
$$Q4:2\pi-\frac{\pi}{4}=\frac{5\pi}{4}$$
However, this is only for $\tan x=-1$, my professor told me to apply the "modifications of x" after.
$$3x=\frac{3\pi}{4}=\frac{3\pi}{4}\div3=\frac{\pi}{4}$$
$$3x=\frac{5\pi}{4}=\frac{5\pi}{4}\div3=\frac{5\pi}{12}$$
Is this the correct way to solve this equation? I believe dividing out $\sin3x$ leads to domain or range errors. However, I tried solving by adding $\sin3x$ to both sides first but could not figure out how to isolate the trig functions. I will clarify anything needed in the comments.

HINT
I would recommend you to proceed as follows:
\begin{align*}
\cos(3x) = -\sin(3x) & \Longleftrightarrow \cos(3x) = \sin(-3x)\\\\
& \Longleftrightarrow \cos(3x) = \cos\left(\frac{\pi}{2} + 3x\right)\\\\
& \Longleftrightarrow 3x = -\frac{\pi}{2} - 3x + 2k\pi\\\\
& \Longleftrightarrow x = -\frac{\pi}{12} + \frac{k\pi}{3}
\end{align*}
where $k\in\mathbb{Z}$.
Hopefully this helps!
