<s>[INST] 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.
 [/INST] though ur answer is not wrong, as long as u show that $\sin(3x) \neq 0$
a better way to solve would be
$\cos( 3x) + \sin(3x) = 0$
so $\sqrt{2}\sin(3x + \frac{\pi}{4}) = 0$
let y= $3x + \frac{\pi}{4}$
$ y \in [\frac{\pi}{4}, 6\pi+\frac{\pi}{4}]$
$\sin(y) =0$
so $ y \in \{\pi , 2\pi , 3\pi , 4\pi , 5\pi , 6\pi\} $
so $x = \frac{\pi}{4}$ or $\frac{7\pi}{12}$ or $\frac{11\pi}{12}$ or $\frac{5\pi}{4}$ or $\frac{19\pi}{12}$ or $\frac{23\pi}{12}$  or $\frac{9\pi}{4}$
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