What is all the solutions to tan(2x) - tan x=0 and how do you find it
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# What is all the solutions to tan(2x) - tan x=0 and how do you find it

[From: ] [author: ] [Date: 11-07-08] [Hit: ]
(1) tan(2x) = 2 tan x/ [1 - tan^2(x)],(6) For [ tan^2(x) ] / {1 - tan^2(x)} = 0 --->, alsox = 0 and 180 deg.tan x = 0,1 + tan²x = 0 tan²x = –1,if tanx = 1,......
Note that tan (a+b) = (tan a + tan b) / (1 - tan a * tan b). Now if we let a = b = x, then
tan 2x = 2 tan x / (1 - (tan x)^2)

tan 2x - tan x = 0
(2 tan x)/(1 - (tan x)^2) - tan x = 0
(2 tan x)/(1 - (tan x)^2) = tan x
2 tan x = tan x (1 - (tan x)^2)
2 tan x = tan x - (tan x)^3
tan x + (tan x)^3 = 0
tan x (1 + (tan x)^2) = 0

Since 1 + (tan x)^2 =/= 0 for any x we can divide both sides by it.

tan x = 0
x = k pi for any integer k.

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tan(2x) - tan x = 0 --->

(1) tan(2x) = 2 tan x/ [1 - tan^2(x)], an identity

(2) tan(2x) - tan x = 2 tan x/ [1 - tan^2(x)] - tan x = tan x [2 / {1 - tan^2(x)} - 1] = 0

(3) tan x [2 / {1 - tan^2(x)} - 1] = tan x [ 2 - 1 + tan^2(x) - 1] / {1 - tan^2(x)} = 0

(4) tan x [ 2 - 1 + tan^2(x) - 1] / {1 - tan^2(x)} = tan x [ tan^2(x) ] / {1 - tan^2(x)} = 0

(5) For tan x = 0 ---> x = 0 and 180 deg

(6) For [ tan^2(x) ] / {1 - tan^2(x)} = 0 --->, also x = 0 and 180 deg.

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2tan x/(1 – tan²x) – tan x 0= 0

2tan x –tan x + tan³x = tan x(1 + tan²x) = 0

tan x = 0, x = kπ where k is any integer

1 + tan²x = 0 tan²x = –1, no solution

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tan2x = 2tanx/(1-tan^2x)
2tanx = (1-tan^2x) X tanx
2tanx = tanx -tan^3x
tan^3x -tanx = 0
tanx(tan^x -1 ) = 0
tanx = 0
then x= 0
if tanx = 1, then x= 45deg

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tan2x= 2tanx/1-tan^2x
Using this , I'm sure you can solve further!!!

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C[1] is Integers && (x = 2 pi C[1] || x = pi + 2 pi C[1])
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