How do you solve $\sin x = \dfrac{{\sqrt 2 }}{2}$

Hint: Here we will simplify the given equation and then by using the trigonometric table for doing some simplification, we will find the value of $x$.

Complete step-by-step solution:
The given term is: $\sin x = \dfrac{{\sqrt 2 }}{2}$
Now we know that $2 = \sqrt 2 \times \sqrt 2 $, therefore on substituting it with the denominator of the right-hand side we get:
$ \Rightarrow \sin x = \dfrac{{\sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }}$
Now on simplifying we get:
$ \Rightarrow \sin x = \dfrac{1}{{\sqrt 2 }}$
Now from the trigonometric table we know that:
$ \Rightarrow \sin \left( {\dfrac{\pi }{4}} \right) = \left( {\sin \pi - \left( {\dfrac{\pi }{4}} \right)} \right)$
This can be written as:
$ \Rightarrow \sin \left( {\dfrac{{3\pi }}{4}} \right)$
This has the value:
$ \Rightarrow \dfrac{1}{{\sqrt 2 }}$ therefore,
$x = {\left( {\dfrac{\pi }{4}} \right)^c}$ which is ${45^\circ }$, $x = {\left( {\dfrac{{3\pi }}{4}} \right)^c}$ which is ${135^\circ }$
Therefore, on generalizing the answer we get:
$x = \left( {2n\pi + \dfrac{\pi }{4}} \right)or\left( {2n\pi + \dfrac{{3\pi }}{4}} \right),n \to \varepsilon \to \mathbb{Z}$, which is the required answer.

Therefore the value of x from $\sin x = \dfrac{{\sqrt 2 }}{2}$ is equal to $45^\circ$.

Note: This question can also be done by using the inverse trigonometric function as:
We have the given equation after simplification as:
$ \Rightarrow \sin x = \dfrac{1}{{\sqrt 2 }}$
Now using the inverse trigonometric function, we get:
$ \Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$
Therefore, the principal value of $\sin x$ is $\dfrac{\pi }{4}$
Now since sine is positive in the first and second quadrant, we will subtract the principal value from $\pi $to get the solution in the second quadrant.
Therefore,
$ \Rightarrow \pi - \dfrac{\pi }{4}$
On simplifying we get:
$ \Rightarrow \dfrac{{3\pi }}{4}$, which is the solution in the second quadrant.
Therefore, on generalizing the answer we get:
$ \Rightarrow x = \left( {2n\pi + \dfrac{\pi }{4}} \right)or\left( {2n\pi + \dfrac{{3\pi }}{4}} \right),n \to \varepsilon \to \mathbb{Z}$, which is the required answer.
It is to be remembered which trigonometric functions are positive and negative in what quadrants.
The formula used over here is for $\sin (n\pi + x)$ ,
It is to be remembered that $\sin (n\pi + x) = {( - 1)^n}\sin x$
Basic trigonometric formulas should be remembered to solve these types of sums.
The inverse trigonometric function of $\sin x$ which is ${\sin ^{ - 1}}x$ used in this sum
For example, if $\sin x = a$ then $x = {\sin ^{ - 1}}a$
And ${\sin ^{ - 1}}(\sin x) = x$ is a property of the inverse function.
There also exists inverse function for the other trigonometric relations such as tan and cos.
The inverse function is used to find the angle $x$ from the value of the trigonometric relation.