<s>[INST] Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\ 
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 \\ 
\tan ^2 x & = 3
\end{align}
This is as far as I can get, and when I look at the Mark Scheme no other Trignometric Identities have been used. Thanks
 [/INST] Hint. As written, the assertion is not correct since

$$
\tan\left(\frac{2\pi}3\right)+\tan\left(\frac{\pi}3\right)=0
$$ but $$\tan\left(\frac{\pi}3\right)=\sqrt{3}\color{red}{\neq}0.$$

There is a mistake in your reasoning, starting as you did you obtain
$$
 \left(\frac{2\color{blue}{\tan x}}{1-\tan ^2 x}+\color{blue}{\tan x}\right)=0
$$ which one may rewrite as
$$
\color{blue}{\tan x} \times\left(\frac{2}{1-\tan ^2 x}+1\right)=0
$$ or
$$
\tan x \times\left(\frac{\color{red}{3-\tan^2 x}}{1-\tan ^2 x}\right)=0
$$ giving 

$$
\tan x=0 \quad \text{or}\quad\color{red}{\tan^2 x=3}.
$$ 

that is explicitly
$$
x=\pm k\pi \quad \text{or}\quad x=\pm \dfrac{\pi}3\pm k\pi
$$ $k=0,1,2,\ldots$.
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