Solving for $c$: $a = \frac {b - \cos(c)}{\cos^2(c)}$

Question

I have a function that I cannot seem to solve and I really need help please :(

Given $a$ and $b$:

$a = \dfrac{b-\cos(c)}{\cos^2(c)}$ with $0 \leq c < \dfrac{\pi}{2}$

I can graph this if I just provide $b$ and use $c$ as the input variables, but I need to find a discrete answer, or answers.

Any help would be so very much appreciated, as I'm entirely stuck :)

Answer 1

Hint: Multiply both sides by $\cos^2c$, and rearrange a bit, and you have yourself a quadratic equation with $\cos c$ being the unknown.

Answer 2

Hint: set $\cos(c)=x$ and obtain a quadratic equation for $x$. Solve for $x$ and then use $\arccos$ to find an expression for $c$.

Answer 3

I guess you'll get a discussion along these lines $$a=\frac{b-\cos c}{\cos^2c}\iff \begin{cases}\cos c\ne 0\\ \frac{b}{\cos^2c}-\frac1{\cos c}-a=0\end{cases}\iff\\\iff\begin{cases}\cos c\ne 0\\ \frac1{\cos c}=\frac{1-\sqrt{1+4ab}}{2b}\\ b\ne 0\end{cases}\lor\begin{cases}\cos c\ne 0\\ \frac1{\cos c}=\frac{1+\sqrt{1+4ab}}{2b}\\ b\ne 0\end{cases}\lor\begin{cases}\cos c\ne 0\\ b=0\\ \frac1{\cos c}=-a\end{cases}\stackrel{\text{since }c\in\left[0,\frac\pi2\right)}\iff\\\iff\begin{cases} \frac{1-\sqrt{1+4ab}}{2b}\ge 1\\ 1+4ab\ge 0\\ b\ne 0\\c=\arccos\frac{2b}{1-\sqrt{1+4ab}}\end{cases}\lor\begin{cases} \frac{1+\sqrt{1+4ab}}{2b}\ge 1\\ 1+4ab\ge 0\\ b\ne 0\\ c=\arccos\frac{2b}{1+\sqrt{1+4ab}}\end{cases}\lor\begin{cases} b=0\\ a\le-1\\ c=\arccos\left(-\frac1a\right)\end{cases}$$

Those additional inequalities you see up there in the systems, albeit important, are rather trite to discuss any further, so I won't do it. You might be interested in the rationalization $\frac{2b}{1\pm\sqrt{1+4ab}}=\frac{-1\pm\sqrt{1+4ab}}{2a}$.

Answer 4

As mentioned, you can substitute a variable for $\cos(c)$, solve the quadratic equation for that variable. Then, plug in $\cos(c)$ again and solve the final equation using arccosine. $$a = \frac{b-\cos(c)}{\cos^2(c)}$$ $$a\cos^2(c) = b-\cos(c) \implies a\cos^2(c)+\cos(c)-b=0$$ $$x = \cos(c)$$ $$ax^2+x-b=0$$ $$\implies x = \frac{-1\pm\sqrt{1^2-4(a)(-b)}}{2a}$$ $$\implies x = \frac{-1\pm\sqrt{1+4ab}}{2a}$$

Now that we’ve solved for $x$, we can go back to $x = \cos(c)$.
$$\implies c = \cos^{-1}(x) \implies c = \cos^{-1} \left(\frac{-1\pm\sqrt{1+4ab}}{2a}\right)$$