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
 A: 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
$$
x=\pm k\pi \quad \text{or}\quad x=\pm \dfrac{\pi}3\pm k\pi
$$ $k=0,1,2,\ldots$.
A: Notice that the following conditions are equivalent:
\begin{align*}
\tan x + \tan y &= 0\\
\tan x &= -\tan y\\
\tan x &= \tan(-y)\\
x &= -y + k\pi\\
x+y&= k\pi
\end{align*}
If we use $y=2x$ we get
\begin{align*}
3x &= k\pi\\
 x &= k\frac\pi3
\end{align*}
A: In line 4 you divided both side with $\tan x$ assuming that $\tan x\neq0$
($1$) So, if $\tan x\neq0$, you are on right track. $x=60^\circ$
($2$) But, if $\tan x=0$, you didn't consider this one. So, $\tan x=0\implies x=0^\circ$
A: $$\tan x+\tan2x=\dfrac{\sin(x+2x)}{\cos x\cos2x}$$
So, we need $\sin3x=0\implies3x=n\pi$ where $n$ is any integer 
A: By the double angle formula we get $$\tan(2x)+\tan(x)=\frac{2\tan(x)}{1-\tan^2(x)}+\tan(x)=\frac{3-\tan^2(x)}{1-\tan^2(x)}\tan(x),$$ so that $\tan(x)=0$ is certainly a solution.
But $\tan(x)=\pm\sqrt3$ as well, so that the initial claim is false.
