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kikij22

  • 3 years ago

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

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  1. ZeHanz
    • 3 years ago
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    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)

  2. kikij22
    • 3 years ago
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    How do you find all of the answers

  3. kikij22
    • 3 years ago
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    @ZeHanz

  4. ZeHanz
    • 3 years ago
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    Hold on, I'm thinking ;)

  5. ZeHanz
    • 3 years ago
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    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...

  6. ZeHanz
    • 3 years ago
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    (divided it by 7 btw)

  7. ZeHanz
    • 3 years ago
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    Simplify this fraction-mess:\[\frac{ \frac{ \tan x(1-\tan^2x)+2\tan x }{ 1-\tan^2x } }{ \frac{ 1-\tan^2x-2\tan^2x }{ 1-\tan^2x } }-3\tan x=0\]Multiply numerator and denomminator by 1-tan²x:\[\frac{ 3\tan x-\tan^3x }{ 1-3\tan^2x }=3\tan x\]Multiply LHS and RHS by 1-3tan²x:\[3\tan x-\tan^3x=3\tan x-9\tan^3x \Leftrightarrow 8\tan^2x=0\]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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