Find all the solutions for \[\sin x = a\]?

Hint: The inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios. We will convert it into \[x = si{n^{ - 1}}\left( a \right)\] and then we will find the solutions for this equation. For all natural numbers, either odd or even, we will find the solutions and then combine in one solution to get the final output.

Complete step by step solution:
Given that,
\[\sin x = a\]
Here, $x$ could be either in radians or in degrees.
Let take $x$ to be in radians. Then, one solution of this equation would be \[x = si{n^{ - 1}}\left( a \right)\] as $sin^{-1}$ is the inverse trigonometric function for $\sin$.
As we know, adding 2π to an angle gives an angle equivalent to the above value of x.
This means that we could keep adding 2π to this angle and get infinitely more solutions.
Thus, \[x = si{n^{ - 1}}\left( a \right)\] is the only solution for this.

Suppose, \[A = si{n^{ - 1}}\left( a \right)\].
Thus, a more complete solution would be \[x = A + \left( {2\pi } \right)n\] for all \[n \in N\].
Here, multiplying by ‘n’ means that adding any multiple of 2π is a solution.
In short, all complementary angles are also a solution. This means that the sum of complementary angles is equal to \[{180^ \circ }\] or \[2\pi \].
Thus, for every solution of ‘A’ another solution is: \[\pi - A\]
As for solving this,
\[ = A + (\pi - A)\]
\[ = \pi \]
As before, to this other solution too, we can add multiples of \[2\pi \] to have equivalent solutions.
Thus, another set of solutions is: \[x = \pi - A + \left( {2\pi } \right)n\]
Hence, a complete solution for \[\sin x = a\] is:
1) \[x = A + \left( {2\pi } \right)n\]
2) \[x = \pi - A + \left( {2\pi } \right)n\]
Next, we will rearrange both the equations as below:
For the first equation:
\[ \Rightarrow x = A + \left( {2n} \right)\pi \]
Since, \[n \in N\] and so \[\left( {2n} \right)\] is always an even number. This means, for $n$ is even, we will use this solution.
Then, for the second equation:
\[ \Rightarrow x = - A + 2n\pi + \pi \]
\[ \Rightarrow x = - A + (2n + 1)\pi \]
Since, \[n \in N\] and so \[(2n + 1)\] is always an odd number. This means, for $n$ is odd, we will use this solution.
Thus, we will combine both the rearranged equations and write the two solutions as:
\[{( - 1)^n}A + n\pi \] .
Hence, all the solutions for \[\sin x = a\] is \[{( - 1)^n}A + n\pi \] where \[A = si{n^{ - 1}}\left( a \right)\] for all \[n \in N\].

Note:
Inverse trigonometric functions are also called Arc Functions since, for a given value of trigonometric functions, Because they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent.