Solving for $x$ in $\tan(3x) \tan (2x)= 1$

Question

If $$\tan(3x) \tan(2x)= 1$$ Then $x$ is equal to

Attempt: I used the '$\tan$' identity but it showed no results.

The identity:

$$\frac{\tan(2x)+\tan(3x)}{1-\tan(2x)\tan(3x)}$$

Answer 1

Hint: $$\tan 3x\tan 2x=1$$ $$\implies \sin 3x \sin 2x= \cos 3x \cos 2x$$ $$\implies \cos 3x\cos 2x-\sin 3x\sin 2x=0$$ Can you see the $\cos(\alpha+\beta)$ identity?

Answer 2

Here's another way:

The equation is the same as $$\tan3x=\cot 2x=\tan(90-2x)$$ $$\Rightarrow 3x=90-2x+n.180$$ $$\Rightarrow x=18+n.36$$

Answer 3

Is this correct ?

$1 - \tan{2x}\tan{3x}=0$

$\therefore\frac{\tan{2x} + \tan{3x}}{1 - \tan{2x}\tan{3x}}= \text{undefined}$

$\therefore\tan{5x}=\text{undefined}$

$\therefore\tan{5x}=\tan{\frac{\pi}2}\;\text{or}\;\tan{\frac{3\pi}2}$

$\therefore 5x=90^\circ\;\text{or}\;270 ^\circ$

$\therefore x=18^\circ\;\text{or}\;54 ^\circ$