I"m perplexed by the $2\theta.$
I diverted $r^2$ to gain $r^2 = \frac1\cos2\theta$
Now, usually if that was just a $\cos \theta$ I would multiply both political parties by $\frac1r$ and then substitute $r$ for $\sqrt x^2+y^2$ and then instead of $r\cos\theta$ because that $x$. But the $2\theta$ doesn"t allow for the to happen.
You are watching: Find a cartesian equation for the curve and identify it. r2 cos(2θ) = 1
Since $\cos 2 \theta = 2 \cos^2 \theta -1$, then $$r^2 \cos 2 \theta = 2 r^2 \cos^2 \theta - r^2 = 2(r \cos \theta)^2 - r^2 = 2x^2 - (x^2 + y^2) = x^2 - y^2 ,$$ so her curve is $x^2 - y^2 = 1$ which is recognized to it is in a hyperbola having actually the lines $y = \pm x$ together asymptotes.
Using $cos2\theta$ = $cos^2\theta - sin^2\theta$,
$r^2(cos^2\theta - sin^2\theta$) =1
$(rcos\theta)^2$ - ($rsin\theta)^2$ = 1
$x^2 - y^2 = 1$.
This is a hyperbola centred in ~ the origin.
There space a couple of ways come approach, however here"s a hint for just how I"d do it: $\cos 2\theta = 2\cos^2\theta - 1$, and also $x = r \cos\theta$, therefore $x^2 = \dots$, and so...
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