Find the solution for given trigonometric equation $\dfrac{\tan 3x\text{ }-\text{ tan2x}}{1\text{ + tan3}x\ \text{tan2}x}\text{ = 1}$

Hint: Use the formula for compound angles $\text{tan}\left( \text{A }-\text{ B} \right)\text{ = }\dfrac{\text{tanA }-\text{ tanB}}{1\text{ + tanA}\ \text{tanB}}$ to simplify the left hand side of the given equation. Then find the general solution for the simplified equation, also mention the principal values for the equation.

Complete step-by-step solution -
It is given that, $\dfrac{\tan 3x\text{ }-\text{ tan2x}}{1\text{ + tan3}x\ \text{tan2}x}\text{ = 1}$ .... (i)
Now we know the formula for tan of difference of angles, in case of compound angles,
     $\text{tan}\left( \text{A }-\text{ B} \right)\text{ = }\dfrac{\text{tanA }-\text{ tanB}}{1\text{ + tanA}\ \text{tanB}}$
Comparing the formula with the left hand side of the equation (i), we can put A = 3x and B = 2x.
Thus, putting A = 3x and B = 2x in the formula, we get,
$\begin{align}
  & \dfrac{\tan 3x\text{ }-\text{ tan2}x}{1\text{ + tan3}x\ \text{tan2}x}\text{ = tan}\left( 3x-2x \right) \\
 & \text{ = }\tan x \\
\end{align}$
Therefore, we can say,
$\begin{align}
  & \dfrac{\tan 3x\text{ }-\text{ tan2x}}{1\text{ + tan3}x\ \text{tan2}x}\text{ = 1} \\
 & \therefore \text{ }\tan x\text{ = 1} \\
\end{align}$
We know that $\tan \dfrac{\pi }{4}\text{ = 1 }$. Thus, the principal value for x when tanx = 1 is $x\text{ = }\dfrac{\pi }{4}$.
But we know, tanx is a periodic function of x. Thus, it repeats its value for different values of x after a certain interval, for a countably infinite number of points corresponding to x.
As tanx = 1, that is, a positive value, the angle x necessarily belongs to the first and third quadrant.
Thus, the general solution for x is given by,
$\begin{align}
  & \tan x\text{ = 1} \\
 & \therefore \text{ }x\text{ = n}\pi \text{ + }\dfrac{\pi }{4},\text{ n = 0, }\pm \text{1, }\pm \text{2,}...... \\
\end{align}$
Thus, for all the integer values of n, the general solution of x for the given equation is x = $\text{n}\pi \text{ + }\dfrac{\pi }{4}$.
The principal value of x is given by x = $\dfrac{\pi }{4}$, when n = 1.

Note: In a question, if one is asked to find out the solution for a trigonometric expression, then we have to find out the general solution for the angles, since trigonometric functions are generally periodic in nature. It is better to also mention the principal value of the angle, along with the general solutions.