Make quadratic equation in terms of $\cos^2x$ if given $\sin x=\dfrac{\sqrt5 -1}{2}$

Question

I have $\sin x=\dfrac{\sqrt5 -1}{2}$

And I need to make a quadratic equation in terms of $\cos^2 x$

$\cos^4 x +\cos^2 x =1$

My attempt:

$\sin^2 x=\dfrac{6-2\sqrt 5}{4}$ but by plugging $\sin^2x$ in $\sin^2x +\cos^2 x=1$ it only giving some numerical value of $\cos^2x$ not quadratic equation.

Thanks in advance.

Answer 1

Hint: $\cos^2 x=\frac{\sqrt{5}-1}{2}=\sin x \implies \cos ^4 x=\sin^2 x=1-\cos^2 x$

Answer 2

$$\sin x = \frac{\sqrt{5}-1}{2}\implies \sin^2x=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$$

$$\cos^2x = 1-\frac{3-\sqrt{5}}{2}=\frac{\sqrt{5}-1}{2}\implies \cos^4x=\frac{3-\sqrt{5}}{2}$$

Then $$\cos^4x+\cos^2x = \frac{3-\sqrt{5}+\sqrt{5}-1}{2}=\frac{2}{2}=1$$

as required.