Cubic equation with circle intersection to form a square 
A cubic equation and circle (unit radius) has intersection at A,B,C,D. ABCD is a square. Find the angle $\theta$.
I tried:

*

*$(0,0)$ is a solution so constant term is $0$


*Substituting A(x,y) and C(-x,-y) and adding them gives coefficient of $x^2$ is 0.
Then the cubic becomes f(x) = $ax^3+bx$.
3.Substituting A and B and added the two equations.
I found it interesting-for n points given we can find a unique n+1 degree polynomial
Also - Can complex number be used here?
Please note: I am not sure whether we can find the angle(integer) without knowing the coefficients of the cubic.
EDIT: From the answers
1.putting A $(cos\theta,sin\theta)$ in f(x) :
$acos^3\theta + b cos\theta = sin\theta$
2.putting B $(-sin\theta,cos\theta)$ in f'(x) :
$asin^2\theta + b  = tan\theta$ [ as circle has $tan\theta$ slope at B]
$1,2 $ eqn gives $3asin^2\theta = acos^2\theta$
So, $sin^2\theta = \frac{1}{4}$
But I getting the value of $\theta$ but a answer shows plot of many cubics
-> because in my case $ABCD$  is a square.
 A: Let
$$f(x)=ax^3+bx. \quad \quad (1)$$
Let's find out the values of $a$ and $b$ for a specific value of $R$.
Let the points $ A=(R \cos \theta, R \sin \theta)$ and $B=(-R \sin \theta, R \cos \theta)$ two intersection points of the circle $\Lambda (O, R)$ with $f(x)$, so that B is a tangent point.
Substituting the coordinates of $A$ in $f(x)$, we get:
$$\tan \theta =aR^2\cos^2\theta+b. \quad \quad (2)$$
Substituting the coordinates of $A$ and $B$ in $f(x)$, we get with some algebra:
$$a =\frac{4}{R^2 \sin4\theta}. \quad \quad (3)$$
Substituting the $x$ of $B$ in $f'(x)$, which is equal to $\tan \theta$,  we get:
$$\tan \theta =3aR^2 \sin^2\theta +b. \quad \quad (4)$$
Choosing $R =2$ and substituting in $(2)$ and $(4)$, we get:
$$\theta = 30°.$$
The values of $a$ and $b$ are:
$$a =\frac{2}{\sqrt 3}$$
and
$$b =-\frac{5\sqrt 3}{3}.$$
Plot:

A: HINT
The cubic must clearly be of type
$$
y = bx\left( {x^{\,2}  - a^{\,2} } \right)
$$
In polar coordinates
$$
r\sin \theta  = br\cos \theta \left( {r^{\,2} \cos ^{\,2} \theta  - a^{\,2} } \right)
$$
i.e.
$$
0 = r\left( {br^{\,2} \cos ^{\,3} \theta  - \left( {a^{\,2} b\cos \theta  + \sin \theta } \right)} \right)
$$
and excluding the origin
$$
0 = br^{\,2} \cos ^{\,3} \theta  - c\cos \left( {\theta  + \beta } \right)
$$
where either $b$ and $c$ can be taken as positive.
So
$$
r = \sqrt {{{c\cos \left( {\theta  + \beta } \right)} \over {b\cos ^{\,3} \theta }}} 
$$
Then $D$ is a local max for $r$, and you shall impose to find the same $r_{max}$ at $90^{\circ}$ thereafter.
