how to solve the equation $\tan^2(2x)+2\tan(2x) \cdot \tan(3x)-1=0$ How to solve the equation $$\tan^2(2x)+2\tan(2x) \cdot \tan(3x)-1=0$$
Can anyone give some hints in this question ?

Let $t$ = tan $x$.  Using trig, we can see that tan $2x = \frac{2t}{1-t^2},$ and tan $3x = t\frac{3-t^2}{1-3t^2}$.
Then the equation becomes:
$$\frac{4t^2}{(1-t^2)^2} + 2\frac{2t^2(3-t^2)}{(1-t^2)(1-3t^2)} - 1 = 0$$
Multiply by the GCD, which is $(1-t^2)^2(1-3t^2)$ to get
$$(4t^2)(1-3t^2) + 4t^2(3-t^2)(1-t^2) - (1-t^2)^2(1-3t^2) = 0$$
Expand, and solve for $t$, and then take the arctangent.
