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

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The discussion focuses on solving the equation tan x ⋅ tan 4x = 1, with participants evaluating different approaches. One method involves transforming the equation into cos 4x ⋅ cos x - sin 4x ⋅ sin x = 0, leading to cos 5x = 0, which simplifies the solution process and avoids division by zero issues. Another participant questions the justification for a leap from tan x = cot 4x to 3x = kπ, highlighting the need for clarity in the steps taken. Additionally, there is mention of using Taylor series for approximating solutions near zero, suggesting potential values for x. The conversation emphasizes the importance of thoroughness in mathematical derivations.
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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Likes WWGD and chwala
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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The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.

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