# Given that $\tan 2x+\tan x=0$, show that $\tan x=0$

Given that $\tan 2x+\tan x=0$, show that $\tan x=0$

\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

• You have made a mistake between the third and the fourth equality... the bracket expansion is incorrect. One you correct that you can solve the problem, via an intuitive substitution – b00n heT Jun 7 '16 at 11:42
• @b00n heT I divided both sides by $\tan x$ – Thomas Winkworth Jun 7 '16 at 11:43
• How can you if it is zero? – Qwerty Jun 7 '16 at 11:43
• There is your error. Recall that $xf(x)=0 \iff x=0 \lor f(x)=0$. It is a very common mistake to divide out the $x$, and thus to lose a trivial, but important, solution – b00n heT Jun 7 '16 at 11:43
• Having got as far as $\tan x(3-\tan^2x)=0$ you can conclude that either $\tan x=0$ or $\tan x=\pm\sqrt3$. – almagest Jun 7 '16 at 11:48

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

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

• How did you get to $\tan x \times\left(\frac{2}{1-\tan ^2 x}+1\right)=0$ – Thomas Winkworth Jun 7 '16 at 11:50
• @ThomasWinkworth By factoring $\frac{2\color{red}{\tan x}}{1-\tan ^2 x}+\color{red}{\tan x}=0$. Do you see it? – Olivier Oloa Jun 7 '16 at 11:50
• @ThomasWinkworth bro don't worry about that,you are on right track,but you made mistake on divison by 0 – Hailey Jun 7 '16 at 11:51
• I think it is confusing to say the assertion is not correct. It is partly correct, since $\tan(x)=0$ might be a solution to the equation. However, it might not be as shown by your example. So the assertion to prove would be the assertion as stated at the end of your answer. Nicely worked out btw.. – Tim Huijgens Jun 7 '16 at 11:53
• I get it, basically $\tan x(3-\tan^2x)=0$ – Thomas Winkworth Jun 7 '16 at 13:14

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*}

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, $\tan x=0\implies x=0^\circ$

$$\tan x+\tan2x=\dfrac{\sin(x+2x)}{\cos x\cos2x}$$

So, we need $\sin3x=0\implies3x=n\pi$ where $n$ is any integer

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.