Analytic Geometry/Pre-calc

[kakamilan17]

I'm trying to find the angle of rotation of this equation. x²-3xy+4y²+7=0
I was given x=x'cosθ - y'sinθ and y=x'sinθ+y'cosθ


So I plug in those two equations for x and y and get a really long equation.
I simplified, then I factored out x'y' out of the long equation and was told to set the coefficient of x'y' to 0. So I end with:

cos²θ + 2cosθsinθ - sin²θ = 0

Now I'm stuck trying to use the trig identities to solve for the angle.
I could solve this using logic, but as this problem is for Analytic Geometry, I'm supposed to prove everything using math. So how can I solve for the angle in this case? I can't seem to be able to make headway with the identities.

So which identities could I use to solve this last part?

-Thank you very much

 

[soroban]

Hello, kakamilan17!

\(\displaystyle \cos^2\!\theta + 2\sin\theta\cos\theta - \sin^2\!\theta \:=\:0\)

\(\displaystyle \text{We have: }\:\underbrace{\cos^2\!\theta - \sin^2\!\theta}_{\cos2\theta} + \underbrace{2\sin\theta\cos\theta}_{\sin2\theta} \:=\:0\)

. . . \(\displaystyle \cos2\theta + \sin2\theta \:=\:0 \quad\Rightarrow\quad \sin2\theta \:=\:-\cos2\theta \quad\Rightarrow\quad \dfrac{\sin2\theta}{\cos2\theta} \:=\:-1 \)

. . . \(\displaystyle \tan2\theta \:=\:-1 \quad\Rightarrow\quad 2\theta \:=\:-\dfrac{\pi}{4} \quad\Rightarrow\quad \theta \:=\:-\dfrac{\pi}{8}\)