Question

The general solution of $\tan 3x=1$ is
A) $n\pi +\dfrac{\pi }{4}$
B) $\dfrac{n\pi }{3}+\dfrac{\pi }{12}$
C) $n\pi $
D) $n\pi \pm \dfrac{\pi }{4}$

Answer 1

Hint:
Here, we need to find the general solution of$\tan 3x=1$. For that, we will put the value of 1, which is$\tan \dfrac{\pi }{4}$ , then we will equate $\tan 3x$with$\tan \dfrac{\pi }{4}$. And we will find the general solution of$\tan 3x=\tan \dfrac{\pi }{4}$. As the general solution of $\tan \theta =\tan \alpha $ is$\theta =n\pi +\alpha $. We will replace $\theta$ with 3x and replace $\alpha$ with $\dfrac{\pi }{4}$ and we will find the required general formula.

Complete step by step solution:
It is given:
$\tan 3x=1$, we have to find its general solution.
We know,$\tan \dfrac{\pi }{4}=1$, so we will put $\tan \dfrac{\pi }{4}$ in place of 1.
Therefore,
$\tan 3x=\tan \dfrac{\pi }{4}$
We know the general solution of is.
Here, we will replace $\theta $ with 3x and replace $\alpha $with $\dfrac{\pi }{4}$.
So, the general solution of $\tan 3x=\tan \dfrac{\pi }{4}$is $3x=n\pi +\dfrac{\pi }{4}$
Rewriting the general solution obtained, we get
$3x=n\pi +\dfrac{\pi }{4}$
Now, we will divide both sides by 3.
$\dfrac{3x}{3}=\dfrac{n\pi +\dfrac{\pi }{4}}{3}$
After simplification, we get
$x=\dfrac{n\pi }{3}+\dfrac{\pi }{12}$
Therefore, the general solution of $\tan 3x=1$is$x=\dfrac{n\pi }{3}+\dfrac{\pi }{12}$.

Thus, the correct option is B.

Note:
We have used the general solution of $\tan \theta =\tan \alpha $here. To understand it deeply, let’s look at the derivation of their general solution.
We want to find the general solution of $\tan \theta =\tan \alpha $here.
Rewriting equation, we get
$\tan \theta =\tan \alpha $
We will break $\tan \theta \And \tan \alpha$in terms of sine and cosine now.
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{\sin \alpha }{\cos \alpha }$
Now, we will subtract $\dfrac{\sin \alpha }{\cos \alpha }$from both sides.
$\dfrac{\sin \theta }{\cos \theta }-\dfrac{\sin \alpha }{\cos \alpha }=\dfrac{\sin \alpha }{\cos \alpha }-\dfrac{\sin \alpha }{\cos \alpha }$
On further simplification, we get
$\dfrac{\sin \theta }{\cos \theta }-\dfrac{\sin \alpha }{\cos \alpha }=0$
We will find the difference of these two fractions.
$\dfrac{\sin \theta .\cos \alpha -\sin \alpha .\cos \theta }{\cos \theta .\cos \alpha }=0$
We know the formula:-$\sin (\theta -\alpha )=sin\theta .\cos \alpha -\sin \alpha .\cos \theta $
Therefore,
$\dfrac{\sin \left( \theta -\alpha \right)}{\cos \theta .\cos \alpha }=0$
Fraction would be 0 only when $\sin \left( \theta -\alpha \right)$is 0.
$\sin \left( \theta -\alpha \right)=0$
We know, the general solution of $\sin \theta =0$ is$\theta =n\pi $.
Therefore, the general solution of $\sin \left( \theta -\alpha \right)=0$is $\theta -\alpha =n\pi$
Rewriting the general solution obtained, we get
$\theta -\alpha =n\pi $
Adding on both sides, we get
$\begin{align}
  & \theta -\alpha +\alpha =n\pi +\alpha \\
 & \Rightarrow \theta =n\pi +\alpha \\
\end{align}$
Hence, the general solution of $\tan \theta =\tan \alpha $is$\theta =n\pi +\alpha$.