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.
 A: Since sine and cosine functions can't be zero together, it means in your given equation, they are non-zero. Thus, you can divide by either of them without any loss of generality.
Also, if $x\in[0,2\pi]$ then $3x\in[0,6\pi]$. Thus, you are missing some solutions.
Also, if you are dividing $\cos3x$ by $\sin3x$, you are getting $\cot3x$. If you want $\tan3x$, you need to divide $\sin3x$ by $\cos3x$.
A: 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!
A: Hint
You may use the classic approach for an equation like
$$a\cos x+b\sin x=c.$$
Divide both sides by $\sqrt{a^2+b^2}$ and get
$$\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x=\frac{c}{\sqrt{a^2+b^2}}.$$
Now you can call $\sin y=\frac{a}{\sqrt{a^2+b^2}}$ and $\cos y=\frac{b}{\sqrt{a^2+b^2}}$.
Can you finish?
A: 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 $\frac{9\pi}{4}$
A: Hint-Another Approach

$\text{There is also this: } cos(a)^2+sin(a)^2=1 \Rightarrow sin(a)^2=-cos(a)^2+1$ $\Rightarrow sin(a)= \sqrt{1-cos(a)^2} \text{ (1)}$
$\text{Now by using (1) in the given equation:}$
$\displaystyle cos(3x)=\sqrt{1-cos(3x)^2}\Rightarrow cos(3x)^2=1-cos(3x)^2\Rightarrow 2cos(3x)^2=1\Rightarrow cos(3x)^2=\frac 1 2 \Rightarrow cos(3x)=\pm \sqrt{\frac 1 2}$

It's another approach.Maybe a little bit more difficult but easy to remember.                  Can you continue?
