## kikij22 Group Title Find all solutions in the interval [0, 2π). 7 tan3x - 21 tan x = 0 one year ago one year ago

1. ZeHanz Group Title

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 Group Title

How do you find all of the answers

3. kikij22 Group Title

@ZeHanz

4. ZeHanz Group Title

Hold on, I'm thinking ;)

5. ZeHanz Group Title

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 Group Title

(divided it by 7 btw)

7. ZeHanz Group Title

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.