Write cos2(x) as linear combination of x↦sin⁡(x) and x↦cos⁡(x)

The function ${\displaystyle f\left(x\right)={\cos }^{2}x}$ has ${\displaystyle f(x+\pi )\equiv f\left(x\right)}$, but any linear combination g of ${\displaystyle \cos }$ and ${\displaystyle \sin }$ has ${\displaystyle g(x+\pi )\equiv -g\left(x\right)}$

Say we can do that, then
${\displaystyle {\cos }^{2}x-b\cos x=a\sin x}$
Let ${\displaystyle t=\cos x}$ and square this equation. We get
${\displaystyle t^4-2b{t}^3+b^2{t}^2=a^2-a^2{t}^2}$
which should be valid for all ${\displaystyle t\in [-1,1]}$ and there for for all t. So the polynomials must be equal for all t, so by comparing the coeficients we get 1=0. A contradiction.
So you can not express ${\displaystyle {\cos }^{2}x}$ as linear combination of ${\displaystyle \cos x\text{ and }\sin x}$