Question

The general solution of the equation $\sin 3x = 0$, is
A) $x = 3n\pi ,n \in I$
B) $x = \dfrac{{n\pi }}{3},n \in I$
C) $x = \left( {2n + 1} \right)\dfrac{\pi }{6},n \in I$
D) None of these

Answer 1

As we can see that the given equation is a trigonometric equation, we will use the formula of trigonometric equations to solve this question. First, we will write the right-hand side of the given equation in terms of sine function then with the help of a formula for trigonometric equations we will find the general value of $x$.

Formula used:
If $\sin \theta = \sin \alpha$, then
$\theta = n\pi + {\left( { - 1} \right)^n}\alpha$, here, $n \in I$

Complete step by step answer:
Given, $\sin 3x = 0$
We know that $\sin {0^ \circ } = 0$. So, we can replace the zero in the right-hand side of the above equation by using this trigonometric value. So, on replacing we get,
$\Rightarrow \sin 3x = \sin {0^ \circ }$
Now from the formula for trigonometric equations we know that when $\sin \theta = \sin \alpha$, then
$\theta = n\pi + {\left( { - 1} \right)^n}\alpha$ and $n \in I$
Here,
$\theta = 3x$
$\alpha = 0$
So, now we get,
$\Rightarrow 3x = n\pi + {\left( { - 1} \right)^n}0$
We can see that the second term in the right-hand side will become zero. So, we get,
$\Rightarrow 3x = n\pi$
On further simplification we get:
$\Rightarrow x = \dfrac{{n\pi }}{3}$ and $n \in I$
Therefore, option (B) is correct.

Note:
Here, one important point to note is that $n \in I$ and we have to write it in our solution.
We can also solve this question more specifically by writing the general solution for the trigonometric equation $\sin \theta = 0$.
The general solution for the trigonometric equation $\sin \theta = 0$ is given by;
$\theta = n\pi ,n \in I$
Let’s solve the question by using this specific result. We have,
$\sin 3x = 0$
Now we will use the formula that when $\sin \theta = 0$, $\theta = n\pi ,n \in I$.
Here, we have $\theta = 3x$, so we get,
$\Rightarrow 3x = n\pi$
$\Rightarrow x = \dfrac{{n\pi }}{3},n \in I$
One thing we should note here is that $\theta = n\pi ,n \in I$, is applicable only when $\sin \theta = 0$.
But $\theta = n\pi + {\left( { - 1} \right)^n}\alpha$, is always applicable whenever $\sin \theta = \sin \alpha$.
We can say that $\sin \theta = 0$ is a special case of $\sin \theta = \sin \alpha$ with $\alpha = 0$.