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
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Originally Posted by tomoshaw 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 $\displaystyle 3x^2-xy-4y^2~?$
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.
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.
i see it factorises to (3x-4y)(x+y) but i thought you could only have 1 trigonometric term in each factor?
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.
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