Question

In (0, π), the number of solutions of the equation ​$\tan x+\tan 2x+\tan 3x=\tan x\tan 2x\tan 3x$ is
(a) 7
(b) 5
(c) 4
(d) 2.

Answer 1

(d) 2
Given equation:

$$\begin{gathered} \tan x+\tan 2x+\tan 3x=\tan x\tan 2x\tan 3x \\ \Rightarrow \tan x+\tan 2x=-\tan 3x+\tan x\tan 2x\tan 3x \\ \Rightarrow \tan x+\tan 2x=-\tan 3x(1-\tan x\tan 2x) \\ \Rightarrow \frac{\tan x+\tan 2x}{1-\tan x\tan 2x}=-\tan 3x \\ \Rightarrow \tan (x+2x)=-\tan 3x \\ \Rightarrow \tan 3x=-\tan 3x \\ \Rightarrow 2\tan 3x=0 \\ \Rightarrow \tan 3x=0 \\ \Rightarrow 3x=n\mathrm{\pi } \\ \Rightarrow x=\frac{n\mathrm{\pi }}{3} \end{gathered}$$

Now,
$x=\frac{\mathrm{\pi }}{3},n=1$
$x=\frac{2\mathrm{\pi }}{3},n=2$
$x=\frac{3\mathrm{\pi }}{3}=180°$, which is not possible, as it is not in the interval $(0,2\mathrm{\pi })$.

Hence, the number of solutions of the given equation is 2.