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VCE Specialist 3/4 Question Thread!

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Bhootnike:

--- Quote from: b^3 on December 05, 2011, 08:05:32 pm ---
--- Quote from: Bhootnike on December 05, 2011, 07:56:35 pm ---Find x such that:

4 tan x = 6 tan 2x, 0 ≤ x ≤ 2π

----------------------------------------

I am stuck on this!

4 tan x= 6tan 2x
        =6( 2tan x/1-tan^2 x)
        = 12 tan x/1-tan^2 x
4 tan x(1-tan^2 x) = 12 tan x
1-tan^2 x = 3
3+tan^2 x = 1
root3 +tan x = 1



--- End quote ---
Use the tan double angle formula then see where you can get to.
tan(2x)=(2tan(x))/(1-tan^2(x))

--- End quote ---

:\ i suck haha

b^3:
move everytihng to the one side then used null factor law.

4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, or   or no solution

So x=0, ,

EDIT: forgot sols

pi:
x also equals pi and 2pi from b^3's working

b^3:

--- Quote from: pi on December 05, 2011, 08:14:35 pm ---x also equals pi and 2pi from b^3's working

--- End quote ---
Yeh I'm a bit rusty, I clicked post, then realised and my com froze.

Moderator action: removed real name, sorry for the inconvenience

TrueTears:
yeah other mistake was

4 tan x(1-tan^2 x) = 12 tan x

cancelling the tan killed off solns

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