homogenous eqns (1 Viewer)

[shsshs]

Hi just one query

To solve sin2x + cos2x = (sinx)^2 + 1

the book teaches to expand to
-------------------2--------2---------2---------2----------2
2sinxcosx + cos x - sin x = sin x + sin x + cos x
--------------------2
2sinxcosx - 3sin x = 0

---------------------------2
then DIVIDING by cos x to get
------2
3 tan x - 2 tan x = 0


MY question is.. instead of dividing by (cosx)^2.. why dont u factorise it outside. ie

-------2-----------------2
(3 tan x - 2 tan x)cos x = 0

i cant see whats wrong with this... yet when i solve cosx = 0 .. the answer dusnt work in the original eqn..

help thanks!

 

[pLuvia]

I don't see what you mean, and I think you meant
3tan²x-2tanx=0

What do you mean by factorising? If you were going to factorise it you would end up with
sinx(2cosx-3sinx)=0
Which is the same as
3tan²x-2tanx=0

 

[shsshs]

I don't see what you mean, and I think you meant
3tan²x-2tanx=0

What do you mean by factorising? If you were going to factorise it you would end up with
sinx(2cosx-3sinx)=0
Which is the same as
3tan²x-2tanx=0

yeh i mean from the eqn

--------------------2
2sinxcosx - 3sin x = 0
-----------------------------------------2
how can u just divide that by cos x

shouldnt u factorise it out so that

--------------------2--------2--------2
2sinxcosx - 3sin x = cos x( 3 tan x - 2 tan x) = 0

and that wld giv u an extra solution of cosx = 0 rahter than just the quadratic involving tanx

 

[pLuvia]

If you divide by the LHS you have to divide by the RHS therefore the LHS=RHS eg
2sinxcosx - 3sin²x=0
(2sinxcosx - 3sin²x)/cos²x=0/cos²x
2tanx-3tan²x=0

But you're not dividing the whole equation by cos²x but each term i.e.
(2sinxcosx)/cos²x-(3sin²x)/cos²x=0

That's why that extra cosx=0 isn't required in this equation

 

[pLuvia]

Or you could look at it this way
2sinxcosx-3sin²=0
sinx(2cosx-3sinx)=0
sinx=0 or 2cosx-3sinx=0
x=0,pi,2pi,3pi..
or
2cosx=3sinx
tanx=2/3
x=tan^-12/3

 

[shsshs]

hi still not quite 100% on this one.. altho ur second method was good thanks

how is dividing each term , different to dividing both sides?

if u have

x^2 - x = 0 then you go x(x-1) = 0 so x = 0,1

hows this different from

--------------------2
2sinxcosx - 3sin x = 0
then u go
----2---------2
cos x( 3 tan x - 2 tan x) = 0

ur help has been much appreciated btw

 

[Riviet]

Essentially, you are creating extra solutions that don't work. I have never seen textbook examples ever factorise like that, and I advise you to just factorise with whatever you have.

 

[pLuvia]

But that cos²x isn't a term from the original equation. If you were going to factorise the new equation it would be
1/cos²x²(2sinxcosx-3sin²x)=0
And 1/cos²x=0
There is no solution so your just left with
(2sinxcosx-3sin²x)=0

Or when you expand it you get
2tanx-3tan²x=0

It's hard to explain hope that cleared up a bit

 

[Riviet]

Ah... that explains things, there had to be some mathematical way to explain it. Neat.