Question: Let $\mathscr A$ be a unital C\*-algebra and let $a,b\in \mathscr A$ such that $0\leq a \leq b$ and $a$ is invertible. How to show that $b$ is invertible? 

($0\leq a \leq b$ means that $a,b$ is positive and that $b-a$ is positive. Moreover, a positive element is a hermitian element with a spectrum which is a subset of $[0,\infty)$.)

Since $a$ is invertible, we have that $0 \not \in \sigma(a)$. I guess my question is why this implies that $0 \not \in \sigma(b)$. Maybe one should use functional calculus in some way(?).

Answer: 
Let $x=-\cos(\theta)$ so that

$$\mu(1-x^2)^2-x=0$$

This polynomial indeed has a single root in the range $[0,1]$ for all positive $\mu$.

We can rewrite the above equation as

$$\mu=\frac x{(1-x^2)^2}.$$

For small $x$, $x\approx\mu$.

For $x$ close to $1$, let $1-\epsilon$, we have

$$\mu=\frac{1-\epsilon}{(1-(1-\epsilon)^2)^2}\approx\frac1{4\epsilon^2}.$$

This gives us the approximation

$$x=1-\frac1{2\sqrt\mu}.$$

You can refine the root numerically in the range

$$[\mu,1-\frac1{2\sqrt\mu}].$$

For a "manual" method, you can plot the relation $\mu=f(x)$ as accurately as possible. Then for a given value of $\mu$, find the corresponding $x$ on the plot and use it for a starting value of Newton's iterations.

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**Update**:

The $\mu$ curve has a vertical asymptote at $x=1$, which makes it more difficult to handle. We can discard it by considering the function

$$\frac\mu{\mu+1}=\frac x{(1-x^2)^2\left(\frac x{(1-x^2)^2}+1\right)}=\frac x{(1-x^2)^2+x}.$$

[![enter image description here](https://i.stack.imgur.com/5cxPH.png)](https://i.stack.imgur.com/5cxPH.png)

It turns out that the function is fairly well approximated by $x$ in the range of interest, so that a good initial approximation is simply

$$x=\frac\mu{\mu+1} !$$
You can avoid ugly analytical formulas but you'll need to use a numerical method to solve for the roots of a polynomial.

Rewrite your equation as

$$\mu\sin^4\theta-\sqrt{1-\sin^2\theta}=0$$

(because $\pi/2<\theta<\pi$), from which you get

$$\mu^2\sin^8\theta+\sin^2\theta - 1=0$$

With $x=\sin\theta$ this is equivalent to

$$x^8+\frac{1}{\mu^2}x^2-\frac{1}{\mu^2}=0\tag{1}$$

Solve for the roots of $(1)$, choose the one which is real-valued and positive, and compute the desired angle as

$$\theta\_0=\pi-\arcsin x\_0$$

(again, because $\pi/2<\theta<\pi$).