How do you find two solutions to $ \csc x = 2 $ ?

Hint: Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. Thus the given function can be converted in the form of sine easily. Signs of a trigonometric function are the same in the two of the four quadrants in the graph that’s why any trigonometric function can have many solutions. Using this information, we can find the two solutions of $ \csc x = 2 $ .

Complete step-by-step answer:
We have to find two solutions to the given equation –
 $
  \csc x = 2 \\
   \Rightarrow \dfrac{1}{{\sin x}} = 2 \\
   \Rightarrow \sin x = \dfrac{1}{2} \;
  $
Now, the sine function is positive in the first and second quadrant, we know that $ \sin 30^\circ = \dfrac{1}{2} $ , so the above equation can have two answers as follows –
 $
  \sin x = \sin 30^\circ ,\,\sin x = \sin (180^\circ - 30^\circ ) \\
   \Rightarrow \sin x = \sin 30^\circ ,\,\sin x = \sin 150^\circ \\
   \Rightarrow x = 30^\circ ,\,x = 150^\circ \;
  $
Hence, the two solutions of $ \csc x = 2 $ are $ 30^\circ \,and\,\,150^\circ $ .
So, the correct answer is “ $ 30^\circ \,and\,\,150^\circ $ ”.

Note: A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant. The trigonometric functions are periodic; that is their value repeats after a specific interval; using this property, we find their value in the other quadrants.