Polar curve geometry with half line

Question

Curve:

For reference, the given answer is $cos(a)= \frac{\sqrt{5}-1}{2}$, but I do not get this with my approach and I cannot see where I go wrong. Let $OA=x=a(1+cos(a))$, so $OB=2x$, then the hypotenuse(dotted line) is $3x$, then using trig ratios, $$cos(a)=\frac{adj}{hyp}=\frac{2a}{3x}=\frac{2a}{3a(1+cos(a)}$$ $$cos(a)=\frac{2}{3+3cos(a)}$$ $$3cos^2(a)+3cos(a)-2=0$$ Then by quadratic formula, $$cos(a)=\frac{-3+\sqrt{33}}{6}$$ (ignoring -ve solution)

Answer 1

Note that

$$OA = a ( 1+\cos \alpha),\>\>\>\>\> OB = 2a\sec\alpha$$

Given that $OB = 2OA$, we have,

$$2a\sec\alpha =2a( 1+\cos \alpha) $$

which leads to,

$$\cos\alpha^2 + \cos\alpha -1=0$$

Solve to obtain,

$$\cos\alpha = \frac {\sqrt5 -1}2$$

Answer 2

Let $\cos \theta =c,$ then $$ \frac{2a}{c} = 2a (1+c)$$ Cross multiply for simplifying, solve quadratic equation in the required domain. $$ c+c^2=1,\quad c^2+c-1=0, \quad c = \frac{\sqrt{5}-1}{4} =\cos \alpha. $$

Answer 3

$OA = a(1+\cos \alpha)\\ OB = 2a\sec \alpha\\ 2OA = OB\\ 2a(1+\cos \alpha) = 2a\sec\alpha$

Multiplying through $\frac {\cos \alpha}{2a}$

$\cos^2 \alpha + \cos \alpha = 1$

And solve for $\cos\alpha$ with the Quadratic equation.