Solve the equation: ##\tan x ⋅\tan 4x = 1##

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SUMMARY

The discussion focuses on solving the equation \(\tan x \cdot \tan 4x = 1\). A proposed method involves transforming the equation into \(\cos 4x \cdot \cos x - \sin 4x \cdot \sin x = 0\), leading to \(\cos(5x) = 0\) and subsequently deriving solutions of the form \(3x = k\pi\), where \(k\) is an integer. The approach is noted for avoiding division by zero, a common issue in tangent equations. Additionally, alternative solutions near \(x = 0\) are discussed, yielding approximate values of \(x = 0.218\) or \(-0.218\).

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chwala
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Homework Statement
Solve the equation ##\tan x ⋅\tan 4x = 1##
Relevant Equations
trigonometry
I saw this link; is the approach here correct?

https://www.google.com/search?q=tan+x+tan+4x+=+1&oq=&gs_lcrp=EgZjaHJvbWUqCQgAECMYJxjqAjIJCAAQIxgnGOoCMgkIARAjGCcY6gIyCQgCECMYJxjqAjIJCAMQIxgnGOoCMgkIBBAjGCcY6gIyCQgFECMYJxjqAjIJCAYQIxgnGOoCMgkIBxAjGCcY6gLSAQkyNTc3ajBqMTWoAgiwAgE&sourceid=chrome&ie=UTF-8#fpstate=ive&vld=cid:7ceabfd6,vid:F4aNmm2QblY,st:0In my approach, i worked with:

...
##\cos 4x ⋅\cos x - \sin4 x⋅\sin x=0##

##\cos(4x + x)=0##

##\cos 5x = 0##

From here the solutions are determined easily...

cheers.
 
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This method has the advantage of not knowingly dividing by zero, which <br /> \tan (5x) = \frac{\tan x + \tan 4x}{1 - \tan x \tan 4x} suffers from.
 
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I propose to write tan x= cot 4x because x cannot be 0. Then one will get:
3x=k Pi
where k integer positive or negative
 
bamboum said:
I propose to write tan x= cot 4x because x cannot be 0. Then one will get:
3x=k Pi
where k integer positive or negative
You have skipped a lot of steps going from ##\tan(x) = \cot(4x)## to ##3x = k\pi##. How do you justify this large leap?
 
Others solutions are related to x close to 0. Then tan x is x+1/3x^3. One obtains 4x^2+8/3x^4=1 giving x=0.218 or -0.218. If Taylor's serie is greatest in order perhaps one will get x close to 0.3
 
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