<s>[INST] Show $C(t)=(2\cos^3(t),\sin(2t),2\sin(t))$ is on the intersection of $x^2+y^2+z^2=4$ and $x^2+y^2=2x$ Given a curve defined as: $$C(t)=(2\cos^2(t),\sin(2t),2\sin(t))$$ for $0\le t\le2\pi$
Show that the curve is on the intersection of the sphere $x^2+y^2+z^2=4$ and  $x^2+y^2=2x$.

The intersection of the two equation satisfies:
$$x=2-\frac{z^2}{2}$$
I think the curve is on the the intersection of the two curves if :
$$2\cos^3(t)=2-\frac{4\sin^2(t)}{2}$$
$$\cos^3(t)=\cos^2(t)$$
For all $0\le t\le2\pi$.
But solving the equation implies this is not true in general.
So what is the main strategy and how should I continue?
 [/INST] Check if it belongs to the first surface
Replacing the curve in the first equation we get
$$x^2 + y^2 + z^2 = 
4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^4(t )+ 4\sin^2(t)\cos^2(t) + 4\sin^2(t)$$
which is
$$4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^2(t)[\cos^2(t) + \sin^2(t)] + 4\sin^2(t) = 4\cos^2(t) + 4\sin^2(t) = 4$$
So $C(t)$ is on the first surface.
Check if it belongs to the second surface
Replacing the curve in the second equation we get
$$x^2 + y^2 = 4\cos^4(t )+ \sin^2(2t) =4\cos^4(t )+ 4\sin^2(t)\cos^2(t) =4\cos^2(t)[\cos^2(t) + \sin^2(t)]=4\cos^2(t)=2x$$
So $C(t)$ is on the second surface.
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