General solution of $\tan (2x)\tan (x)=1$

Question

For the question, $\tan (2x) \tan x=1$, I divided it by $\tan x$, and got the solution as $\frac{(2n+1)\pi}{6}$.

$\tan 2x= \cot x= \tan\left(\frac{\pi}{2}-x\right)$. So, $2x=n\pi+ \frac{\pi}{2}-x$. So, $3x= \frac{(2n+1)\pi}{2}$

But the book solved using the formula of $\tan (2x)$, and got the solution as $\frac{(6n \pm 1)\pi}{6}$. I can see that my solution has odd multiples of $\pi/2$, which should be discarded, but I thought of it only after checking the solution. Also, in a way, it suggests that we can't solve these questions in different ways because that way we might get extra solutions. So, how to ensure which method to follow?

Answer 1

I have used a different approach to solve this and tried to keep everything generalized without any predefined assumptions.

$$\tan (2\theta)\tan (\theta)=1 $$

$$ \Longrightarrow \frac{2\tan^2(\theta)}{1-\tan^2(\theta)} -1 = 0$$

$$ \Longrightarrow \frac{3\tan^2(\theta) -1}{1-\tan^2(\theta)} = 0$$

Solving for $3\tan^2(\theta) -1 = 0 $ we get $\theta = n\pi \pm \frac{\pi}{6}$.

However if $\frac{1}{1-\tan^2(\theta)} = 0 $ then $\tan(\theta)$ must be undefined which surely won't satisfy our original solution.

Thus to conclude the solution is $\theta = n\pi \pm \frac{\pi}{6}$.

Answer 2

You got certain solutions, but you must check whether or not the functions that you are dealing with are actually defined at those points. And $\tan x$ doesn't exist when $x$ is of the form $\frac\pi2+k\pi$ ($k\in\mathbb Z$).

Answer 3

Your way is better, I think:

The domain gives $\cos2x\neq0$,$\cos{x}\neq0$ and $\sin{x}\neq0$ and by your idea we obtain: $$\tan2x=\cot{x}$$ or $$\frac{\sin2x\sin{x}-\cos2x\cos{x}}{\cos2x\sin{x}}=0$$ or $$\cos3x=0$$ or $$x=\frac{\pi}{6}+\frac{\pi}{3}k,$$ where $k\in\mathbb Z$.

Now, delete from these roots, roots of $\sin{x}=0$ (which is impossible), $\cos2x=0$ (which is impossible again) and $\cos{x}=0.$

We can get an answer by the following way.

From $\cos3x=0$ on the interval $(0,2\pi)$ we got the following numbers: $$\left\{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\right\}.$$

$\frac{\pi}{2}$ and $\frac{3\pi}{2}$ we need to delete and since $$\frac{11\pi}{6}=\pi+\frac{5\pi}{6}$$ and $$\frac{7\pi}{6}=\pi+\frac{\pi}{6},$$ we got the following series: $$x=\frac{\pi}{6}+\pi k$$ and $$x=\frac{5\pi}{6}+\pi k$$ or $$\left\{\frac{(\pm1+6)\pi}{6}|k\in\mathbb Z\right\}$$

Answer 4

By your method,

$$\tan(2x)=\tan\left(\frac\pi2-x\right)\iff 2x=\frac\pi2-x+k\pi\iff x=\frac{1+2k}{6}\pi$$ and every step is justified.

But the domain of the LHS excludes $$x=\frac{1+2k}{2}\pi$$ and these solutions must be rejected.

By the method of the book,

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

$$2\tan^2x=1-\tan^2x$$ and $$\tan^2x=\dfrac13$$ is only allowed if $\tan^2x\ne1$ and this must be mentioned in the resolution.

Answer 5

$$\tan 2x \tan x=1.~~~(1)$$ $$\Rightarrow \tan 2x = \cot x ~\mbox{if} \tan x \ne 0, \pi/2~~~(2)$$ Therefore, $x \ne k \pi~~~(3)$ and $x \ne (m+1/2)\pi~~~(4).$ From (2), we get $$\tan 2x= \cot x= \tan(\pi/2-x) \Rightarrow 2x =\frac{\pi}{2}-x+n\pi\Rightarrow x= \frac{\pi}{6}+\frac{n\pi}{3}.~~~~(3)$$ These solutions cannot satisfy (3) but they do satisfy (4) when $n=3m+1$ So the complete solution of (1) is given by $$x=\frac{\pi}{6}+\frac{n\pi}{3}, n \in I ~\mbox{but}~ n \ne 3m+1, ~\mbox{where}~ m \in I$$

Answer 6

Writing the equation as $$1=\tan 2x\tan x=\frac{\sin 2x\sin x}{\cos 2x\cos x}\iff\cos 2x\cos x-\sin 2x\sin x=\cos 3x =0$$ with a domain of validity defined by the conditions $\cos x,\cos 2x\ne 0$, i.e. $$x\not\equiv \tfrac\pi2 \mod\pi,\quad x\not\equiv \tfrac\pi 4\mod\tfrac\pi 2,$$ we obtain at once the solutions $$3x\equiv \frac \pi2\mod \pi\iff x\equiv \frac\pi6\mod\frac\pi 3.$$ None of these solutions are congruent to $\frac\pi4\bmod\frac\pi2$, but a part of them are congruent to $\frac\pi2\bmod \pi$. Actually, geometrically the solutions are represented by the vertices of a regular hexagon and the solutions to eliminate are represented by the north and south poles of the trigonometric circle. There remains two series of solutions: $$x\equiv\pm\frac\pi 6\mod \pi.$$