Trig Questions

Using cos^(x/2) = [1 + cos(x)] / 2

Then

2 * [1 + cos(x)] / 2 = cos(2x)  →

1 + cos(x) = cos(2x)  →

And using  cos(2x) = 2cos^2(x) - 1

1 + cosx = 2cos^2(x) - 1    →

2cos^2(x) - cosx - 2 = 0

let x = cosx

2x^2 -x - 2 = 0     .....using the on-site solver......

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{17}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{17}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.780\: \!776\: \!406\: \!404\: \!415\: \!1}}\\
{\mathtt{x}} = {\mathtt{1.280\: \!776\: \!406\: \!404\: \!415\: \!1}}\\
\end{array} \right\}$$

So, using the cosine inverse, and ignoring the second answer, we have.....

cos^-1 [(1 - √17)/4] = x ......  so x = 141.331°  and x = 218.668°

Here is you answer

https://www.desmos.com/calculator/uihaqpesps