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

Hint: 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 \[\alpha = 0\].