How do you solve $2\sin x-1=0$?

Hint: Here we need to solve the given trigonometric equation. For that, we will add or subtract the number from both sides of the equation and then we will divide the number which is multiplied with the variable and then we will use basic trigonometric formulas to get the required value of the variable and hence the required solution of the given trigonometric equation.

Complete step by step solution:
Here we need to solve the given trigonometric equation and the given trigonometric equation is $2\sin x-1=0$.
Now, we will add 1 to both sides of the trigonometric equation.
$\Rightarrow 2\sin x-1+1=0+1$
On further simplifying the terms, we get
$\Rightarrow 2\sin x=1$
Now, we will divide both sides by 2.
$\Rightarrow \dfrac{2\sin x}{2}=\dfrac{1}{2}$
On further simplification, we get
$\Rightarrow \sin x=\dfrac{1}{2}$
We know that the value of $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$. So we will substitute the value here.
$ \Rightarrow \sin x = \sin \dfrac{\pi }{6}$
Now, as we know that the sine function is positive in the first and second quadrant.
Hence, we can write
$x = \dfrac{\pi }{6}$ or $x=\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}$
But we also know that sine function has a cycle of $2\pi $.
Therefore, we can write;
$x=2n\pi +\dfrac{\pi }{6}$ or $x=2n\pi +\dfrac{5\pi }{6}$
We can write it in a single equation as
$\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$
Hence, this is the required solution of the given trigonometric equation.

Note:
Here we have obtained the solution of the given trigonometric equations. Trigonometric equations are defined as the equation which involves the trigonometric functions with angles as the variables. But there is a difference between the trigonometric equation and the trigonometric identities. Trigonometric identities are defined as the equalities which include the trigonometric functions and they are always true for every value of the occurring variables where both sides of the equality are well defined.