Solve the equation $2\cosec\left(x-\dfrac{\pi}{3}\right)=-4$ for $0\le x\le 2\pi$. Solve the equation $\cot(2x+10^\circ)=\tan70^\circ$ for $0^\circ\le x\le 360^\circ$.

1. Convert the cosecant equation into a sine equation.

$2\cosec\left(x-\frac{\pi}{3}\right)=-4$
$\cosec\left(x-\frac{\pi}{3}\right)=-2$
Since $\cosec\theta=\dfrac{1}{\sin\theta}$,
$\sin\left(x-\frac{\pi}{3}\right)=-\frac{1}{2}.$

2. Find the valid interval for the compound angle.

Given
$0\leq x\leq 2\pi,$
subtract $\dfrac{\pi}{3}$ throughout:
$-\frac{\pi}{3}\leq x-\frac{\pi}{3}\leq \frac{5\pi}{3}.$

3. Solve $\sin\theta=-\dfrac12$ in the adjusted interval.

Let
$\theta=x-\frac{\pi}{3}.$
Then
$\sin\theta=-\frac{1}{2}.$
The reference angle is
$\frac{\pi}{6}.$
For $-\frac{\pi}{3}\leq \theta\leq \frac{5\pi}{3},$ the solutions are
$\theta=-\frac{\pi}{6}\quad\text{or}\quad \theta=\frac{7\pi}{6}.$

4. Convert the solutions for the compound angle back to values of $x$.

Since
$x-\frac{\pi}{3}=-\frac{\pi}{6},$
$x=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.$

Since
$x-\frac{\pi}{3}=\frac{7\pi}{6},$
$x=\frac{7\pi}{6}+\frac{\pi}{3}=\frac{9\pi}{6}=\frac{3\pi}{2}.$

1. Rewrite $\tan70^\circ$ as an equivalent cotangent value.

Since $\tan\theta=\cot(90^\circ-\theta)$,
$\tan70^\circ=\cot20^\circ.$
So
$\cot(2x+10^\circ)=\cot20^\circ.$

2. Set up the general solution using the period of cotangent.

Since cotangent has period $180^\circ$,
$2x+10^\circ=20^\circ+180^\circ n,$
where $n$ is an integer.

3. Solve the linear equation for $x$.

$2x=10^\circ+180^\circ n$
$x=5^\circ+90^\circ n.$

4. Find all values of $x$ in the given interval.

For $0^\circ\le x\le360^\circ$,
$x=5^\circ+90^\circ n.$
Taking integer values of $n$:
$n=0,1,2,3$
gives
$x=5^\circ,\ 95^\circ,\ 185^\circ,\ 275^\circ.$