if tan3x-tanx = 3(1+tan3xtanx) what does x= ???

Question

I get as far as tan(2x) = 3
...i'm not sure if i just did it wrong, or i just dont know what to do from there...please help...THANKS

can it be done without using arctan...because i was never taught that...??

Answer 1

I am not sure where you got tan(2x)=3. You first need to write tan(3x) in terms of tanx, use the angle addition identity.

Okay answer below is correct, I was too lazy to look it up.

Answer 2

[tan3x-tanx]/[1+tan3xtanx] = tan(3x-x) = tan(2x), so yes, the original equation is equivalent to tan(2x)=3.

If we assume -pi/2 < 2x < pi/2, then 2x = arctan(3), so x = arctan(3)/2 is one solution.

The set of all solutions is x = arctan(3)/2 + k*pi/2, for any integer k, since then tan(2x) = tan(arctan(3) + k*pi), and tan(t+k*pi) = tan(t) for any integer k and any angle t.

Answer 3

Alternative (much harder!) method:

Convert the equation into:

tan 3x = expression in tan x

Expand tan 3x using addition formula (twice) for tan(A+B) to get

expression in tan x = expression in tan x

Let t = tan x, and you should get

3t^4 + 2t^3 + 2t - 3 = 0

Factorise: ( t² + 1 ) ( 3t² + 2t - 3 ) = 0

etc...... to get x = 0.63 and x = -0.95 (approx) together with appropriate multiples of π/2.

Reconcile with the easy method [ atan(3)/2 ], after showing that

2 atan( (√10 - 1 ) / 3 ) = atan( 3 ).