Find the general solution for cscx = -2

Hint: First we will convert csc into sin and then write that for what value of sin of the angle we get $\dfrac{-1}{2}$, and then we will use the general solution of sin to find all the possible solutions, and we can see that there will be infinitely many solutions of x for which it gives sinx = $\dfrac{-1}{2}$.

Complete step-by-step answer:
Let’s covert csc into sin using the formula $\sin x=\dfrac{1}{\csc x}$
Hence, for cscx = -2 we get $\sin x=\dfrac{-1}{2}$ .
Let’s first find the value of angle for which we get $\dfrac{-1}{2}$.
Now we need to find that in which quadrant sin is negative,
We know that sin is negative in ${3}^{rd}$and ${4}^{th}$ quadrant, so $\pi +\dfrac{\pi }{6}$ and $\dfrac{-\pi }{6}$ both are the correct value,
Here, we will take $\dfrac{-\pi }{6}$.
Now we know that $\sin \left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{2}$
Hence, we get $\sin x=\sin \left( \dfrac{-\pi }{6} \right)$
Now we will use the formula for general solution of sin,
Now, if we have $\sin \theta =\sin \alpha $ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Now using the above formula for $\sin x=\sin \left( \dfrac{-\pi }{6} \right)$ we get,
$x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{-\pi }{6} \right)$
Here n = integer.
Hence, from this we can see that we will get infinitely many solutions for x as we change the value of n.

Note: The formula for finding the general solution of sin is very important and must be kept in mind. In the above solution the value of $\alpha $ we have taken was $\dfrac{-\pi }{6}$, but one can also take the value of $\alpha $ as $\pi +\dfrac{\pi }{6}$ , as it lies in the ${3}^{rd}$quadrant and gives negative value for sin. And then one can use the same formula for the general solution and replace the value of $\alpha $ with $\pi +\dfrac{\pi }{6}$ to get the answer, which is also correct.