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Solving $\tan{2x}=\tan{x}$

Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$

Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$

therefore: $$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$

$$\frac{2\cos x}{2\cos^2x-1} = \frac{1}{\cos x}$$

$$2\cos{^2}(x) = 2\cos{^2}(x) - 1$$

$$0 = -1 ????$$

davidlowryduda
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hondaman
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    You get an absurdity because you canceled of $\sin(x)$ assuming $\sin(x) \neq 0$. –  Mar 27 '13 at 19:06
  • Why do I need to assume $\sin(x) \neq 0$? Aren't the two sides of the equation equal for any value of $ \x $? – hondaman Mar 27 '13 at 19:14
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    $a \times b = a \times c \implies b=c$ only when $a \neq 0$. For instance, $0 \times 2 = 0 \times 3$, doesn't mean $2=3$. Hence, in general if you have $ab=ac$, then either $a=0$ or $b=c$. This means if $b \neq c$, then $a=0$. The two sides are not equal to each other for any value of $x$. –  Mar 27 '13 at 19:16
  • The last line with $\sin(x)$ in it implies one of two possibilities... – Douglas B. Staple Mar 27 '13 at 19:24
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    It makes much more sense now. In hindsight, I should have taken both ratios to one side and factored by $\sin(x)$ which would have given me two condition for the solution. – hondaman Mar 27 '13 at 19:25

3 Answers3

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Recall that $$\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \tan(B)}$$ Hence, $$\tan(2x) = \dfrac{2\tan(x)}{1-\tan^2(x)} = \tan(x)$$ This implies $\tan(x) = 0 \text{ or }1-\tan^2(x) = 2$. $1-\tan^2(x) = 2$ is not possible if $x \in \mathbb{R}$. Hence, $$\tan(x) = 0 \implies x = n \pi$$

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Hint: $$\tan(2x) = \tan(x) \quad \Rightarrow \\ \frac{\sin(2x)}{\cos(2x)} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\ \frac{2\sin(x)\cos(x)}{2\cos^2(x) -1} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\ \frac{2\cos^2(x)}{2\cos^2(x) - 1} = 1 \quad \text{ or }\quad \sin(x) = 0. $$ It looks like you got the "other solution $\sin(x) = 0$. Of course you can't have $0 = -1$. All that means is just that $2\cos^2(x)$ is never equal to $2\cos^2(x) = 1$. So that specific equation does not have any solution. But again, there is the possibility that $\sin(x)= 0$ (which is equivalent to $\tan(x) = 0$.

Jean-Claude Arbaut
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Thomas
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  • I marked this as the correct answer because it uses the same trigonometric identities in my original problem. – hondaman Mar 27 '13 at 19:32
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We know $\tan A=\tan B\implies A=n\pi+B$ where $n$ is any integer (Proof Below )

So, $\tan2x=\tan x\implies 2x=n\pi+x\implies x=n\pi$ where $n$ is any integer

[ Proof:

$\tan A=\tan B\implies \sin A\cos B=\cos A\sin B$

$\implies \sin(A-B)=0\implies A-B=n\pi$ where $n$ is any integer

]

lab bhattacharjee
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