General solution of tan⁡(2x)tan⁡(x)=1

Question

Maritza Mccall · Accepted Answer

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

\tan (2 θ) \tan θ = 1

\Rightarrow \frac{2 \tan^{2} θ}{1 - \tan^{2} θ} - 1 = 0

\Rightarrow \frac{3 \tan^{2} θ - 1}{1 - \tan^{2} θ} = 0

Solving for

3 \tan^{2} (θ) - 1 = 0

we get

θ = n π \pm \frac{π}{6}

However if

\frac{1}{1 - \tan^{2} (θ)} = 0

then

\tan (θ)

must be undefined which surely wont

Eleanor Shaffer · Accepted Answer

tan⁡2xtan⁡x=1     (1)  ⇒tan⁡2x=cot⁡x  if  tan⁡x≠0,π2    (2)  Therefore, x≠kπ   (3) and x≠(m+12)π   (4). From (2), we get  tan⁡2x=cot⁡x=tan⁡(π2−x)⇒2x=π2−x+nπ⇒x=π6+nπ3    (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=π6+nπ3, n∈I  but  n≠3m+1,  where  m∈I

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

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