Solving trigonometrical equations

Question

$\DeclareMathOperator\cosec{cosec}$I want to solve for x $$\cosec2x+\sec426=0$$ I tried $$\cosec(90+2x)=\sec426$$and $$\cosec(90-2x)=\sec426$$ From the question I knew $$\sec426=-\cosec2x$$
I wrote $$\sec426=\cosec(90+426)$$ and $$\sec426=\cosec(90-426)$$ I put 2x as 426 like $90-\theta $ This gave $2x=516$ so, $ \frac{516-360}{2}=78 \,(\text{base})$ I have 102 in quadrant 2 and 282 in quadrant 3
And also $2x=-336$, so $\frac{-336+360}{2}=12$ This gave 348 at quadrant 4 and 168 at quadrant 2. I happened to get the answers but is it possible , and I couldn't understand why the answers have 282 at quadrant 3 , since sine at there is positive. It is also confusing that my second attempt get the values where for sine itself, it is positive and negative. "without using calculator"

Answer 1

$$\csc2x=-\sec426^\circ$$

$$\iff\sin2x=-\cos426^\circ=\sin(90^\circ-426^\circ)$$

$$2x=n180^\circ+(-1)^n(-336^\circ)$$ where $n$ is any integer

If $n$ is even $=2m$(say), $$2x=360^\circ m-336^\circ=24^\circ+360^\circ(m-1)$$

$$x=12^\circ+180^\circ(m-1)\equiv12^\circ{\pmod{180^\circ}}$$

If $n$ is odd $=2m+1$(say)

$$2x=(2m+1)180^\circ+336^\circ$$

$$ x=(2m+1)90^\circ+168^\circ\equiv78^\circ{\pmod{180^\circ}}$$