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ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.0If 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)

kikij22
 one year ago
Best ResponseYou've already chosen the best response.0How do you find all of the answers

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

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.0I 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...

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.0Simplify 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.
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