Trigonometric equation $\tan(x)+\tan(3x)=2\sin(2x)$

Question

Some hint on how to solve this would be very appreciated.

$$ \tan(x)+\tan(3x)=2\sin(2x) $$

Thanks in advance!

Answer 1

Calculation e.g. with:

$\displaystyle \tan x=\frac{\sin x}{\cos x}$

$\displaystyle \tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

$\displaystyle \sin 2x=2\sin x\cos x$

Note:

Using $\enspace\sin^2 x=1-\cos^2 x\enspace$ you will get the following (or equivalent) equations, which are easily to solve:

$\enspace\sin x=0\enspace$ , $\enspace 4\cos^2 x=1\enspace$

$\enspace$(and $\enspace\cos^2 x=1\enspace$ which is solved by $\enspace\sin x=0$)

Hint:

It's senseful to put the solutions into the original equation to test the solutions and to avoid calculation mistakes.

Answer 2

$ \tan x + \tan 3x = 2\sin 2x \\ \implies \frac{\sin x \cos 3x + \cos x \sin 3x}{\cos x \cos 3x} = 2 \sin 2x \\ \implies \frac{\sin 4x}{\cos x \cos 3x} = 2 \sin 2x \\ \implies \frac{2 \sin 2x \cos 2x}{\cos x \cos 3x} = 2\sin 2x \\ $

We can see that $\sin 2x = 0$ is a solution. Therefore, $x = \frac{nπ}{2}$, where $n \in \mathbb{Z}$. For $\sin 2x \neq 0$,

$ \cos 2x = \frac12 \times 2 \cos x \cos 3x \\ \implies \cos 2x = \frac12 \cos 4x + \frac12 \cos 2x \\ \implies \cos 4x = \cos 2x \\ \implies 4x = 2nπ ± 2x \\ \implies x = nπ, \frac{nπ}{3}. $

where $n \in \mathbb{Z}$. Now, going by the initial statement, we need to keep the domain of $\tan \theta$ unchanged. Hence, $x \neq (2k + 1)\frac{π}{2}$ and $x \neq (2k' + 1)\frac{π}{6}$, where $k, k' \in \mathbb{Z}$. But $\sin 2x = 0$ gives us $x = kπ/2$, so the odd multiples of $\frac{π}{2}$ must be eliminated to arrive at the final result. Then again, the even multiples of $\frac{π}{2}$ get included in $nπ$. Therefore, the final result is

$$ x \in\frac{π}{3}\mathbb{Z}$$

Answer 3

By $\displaystyle\tan a+\tan b=\frac{\sin(a+b)}{\cos a\cos b}$ we have $$\frac{\sin(x+3x)}{\cos x\cos 3x}=2\sin2x$$ $$\frac{2\sin(2x)\cos2x}{\cos x\cos 3x}=2\sin2x$$ so $\color{red}{\sin2x=0}$ or $$\frac{2\cos2x}{\cos x\cos 3x}=2$$ $$2\cos2x=\cos4x+\cos2x$$ $$\color{blue}{\cos2x=\cos4x}$$ then there is two answers: $\color{red}{\sin2x=0}$ so $2x=k\pi$ or $\color{red}{x=\dfrac{k\pi}{2}}$, and from $\color{blue}{\cos2x=\cos4x}$ conclude that $4x=2k\pi\pm 2x$ or $\color{blue}{x=k\pi}$ or $\color{blue}{x=\dfrac{k\pi}{3}}$.