How do you solve for $2\sin \left( {3x} \right) - 1 = 0$ ?

Hint: In the given question, we are required to find all the possible values of $\theta $ that satisfy the given trigonometric equation $2\sin \left( {3x} \right) - 1 = 0$ . For solving such type of questions where we have to solve trigonometric equations, we need to have basic knowledge of algebraic rules and identities as well as a strong grip on trigonometric formulae and identities.

Complete step by step answer:
We have to solve the given trigonometric equation $2\sin \left( {3x} \right) - 1 = 0$ .
Taking the constant term to the right side of the equation, we get,
$ \Rightarrow 2\sin \left( {3x} \right) = 1$
$ \Rightarrow \sin \left( {3x} \right) = \left( {\dfrac{1}{2}} \right)$
Now, we can either expand the $\sin \left( {3x} \right)$ term and convert it into expressions consisting of $\sin x$ or write the general solution for the equation $\sin \left( {3x} \right) = \left( {\dfrac{1}{2}} \right)$ and find the value of x directly from there.
So, now we will write the general solution for the equation $\sin \left( {3x} \right) = \left( {\dfrac{1}{2}} \right)$ and carry on with calculations to find the value of x. We know that the value of $\sin \dfrac{\pi }{6}$ is $\left( {\dfrac{1}{2}} \right)$.
We also know that the general solution of the trigonometric equation $\sin \theta = \sin \alpha $ is $\theta = n\pi + {\left( { - 1} \right)^n}\alpha $ where n is any integer. So, we get the general solution of $\sin \left( {3x} \right) = \left( {\dfrac{1}{2}} \right)$ as:
 $ \Rightarrow \left( {3x} \right) = n\pi + {\left( { - 1} \right)^n}{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
$ \Rightarrow \left( {3x} \right) = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{6}} \right)$
Dividing both sides of the equation by $3$, we get,
$ \Rightarrow x = \dfrac{{n\pi }}{3} + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{{18}}} \right)$

So, the solution of the trigonometric equation $2\sin \left( {3x} \right) - 1 = 0$ is $x = \dfrac{{n\pi }}{3} + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{{18}}} \right)$ where n is any integer.

Note: Such trigonometric equations can be solved by various methods by applying suitable trigonometric identities and formulae. The general solution of a given trigonometric solution may differ in form, but actually represents the correct solutions. The different forms of general equations are interconvertible into each other.