factorising trigonometric function

Mar 2013
4
0
Manchester
for -180 </= θ </= 180, solve

3sin^2(θ) - sin(θ)cos(θ) - 4cos^2(θ) = 0

i can't factorise the middle term to separate the sine and cosine
 

Plato

MHF Helper
Aug 2006
22,490
8,653
for -180 </= θ </= 180, solve
3sin^2(θ) - sin(θ)cos(θ) - 4cos^2(θ) = 0
i can't factorise the middle term to separate the sine and cosine

Oh, come on. You can factor \(\displaystyle 3x^2-xy-4y^2~?\)
 
Oct 2012
751
212
Ireland
Let \(\displaystyle sin\theta=x\) and \(\displaystyle cos\theta=y\)

Substitute these into the equation and you get
\(\displaystyle 3x^2-xy-4y^2\)

Edit: Plato beat me to it.
 
Mar 2013
4
0
Manchester
This is how i originally tried to tackle it. I'm either blind or I need to be posting on a key stage 2 board but i just can't see how that factorises and I know it's really simple.
 
Mar 2013
4
0
Manchester
i see it factorises to

(3x-4y)(x+y)

but i thought you could only have 1 trigonometric term in each factor?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
No, you can have whatever is needed in each factor. Now by setting each factor equal to 0, you end up with two simpler equations to solve for theta.