T tomoshaw Mar 2013 4 0 Manchester Mar 18, 2013 #1 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

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

P Plato MHF Helper Aug 2006 22,490 8,653 Mar 18, 2013 #2 tomoshaw said: 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 Click to expand... Oh, come on. You can factor \(\displaystyle 3x^2-xy-4y^2~?\)

tomoshaw said: 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 Click to expand... Oh, come on. You can factor \(\displaystyle 3x^2-xy-4y^2~?\)

S Shakarri Oct 2012 751 212 Ireland Mar 18, 2013 #3 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.

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.

T tomoshaw Mar 2013 4 0 Manchester Mar 18, 2013 #4 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.

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.

T tomoshaw Mar 2013 4 0 Manchester Mar 18, 2013 #5 i see it factorises to (3x-4y)(x+y) but i thought you could only have 1 trigonometric term in each factor?

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 Mar 18, 2013 #6 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.

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.