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Find all solutions in the interval [0, 2π). 7 tan3x  21 tan x = 0
 one year ago
 one year ago
Find all solutions in the interval [0, 2π). 7 tan3x  21 tan x = 0
 one year ago
 one year ago

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ZeHanzBest ResponseYou've already chosen the best response.0
If you write tan3x as tan(x+2x) you can use the formula for tan(a+b)= (tan a + tan a) / (1 − tan a tan b)
 one year ago

kikij22Best ResponseYou've already chosen the best response.0
How do you find all of the answers
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.0
Hold on, I'm thinking ;)
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.0
I first divide the equation by : tan3x  3tanx = 0 Using the sum formula:\[\frac{ \tan x+\tan2x }{ 1\tan x \tan 2x }3\tan x=0\]Now use the formula again, with a=b=x, to get rid of tan2x:\[\frac{ \tan x+\frac{ 2\tan x }{ 1\tan^2x } }{ 1\tan x \frac{ 2\tan x }{ 1\tan^2x } }3\tan x =0\]This is equivalent to\[\frac{ \tan x+\frac{ 2tanx }{ 1\tan^2x } }{ 1\frac{ 2\tan^2x }{ 1\tan^2x } }3\tan x=0\]This is getting a little messy, but we'll keep going on...
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.0
Simplify this fractionmess:\[\frac{ \frac{ \tan x(1\tan^2x)+2\tan x }{ 1\tan^2x } }{ \frac{ 1\tan^2x2\tan^2x }{ 1\tan^2x } }3\tan x=0\]Multiply numerator and denomminator by 1tan²x:\[\frac{ 3\tan x\tan^3x }{ 13\tan^2x }=3\tan x\]Multiply LHS and RHS by 13tan²x:\[3\tan x\tan^3x=3\tan x9\tan^3x \Leftrightarrow 8\tan^2x=0\]Now things are looking sunny again! We have tan x = 0. On [0,2pi) there are only the solutions 0 and pi.
 one year ago
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