Solve a Trigonometric Equation!

Using elementary rules of exponentials write equation as: $2(\color{#3D99F6}{2^{\cos x}})^3-5(\color{#3D99F6}{2^{\cos x}})^2+4(\color{#3D99F6}{2^{\cos x}})-1=0$ This is a trivial cubic equation, substitute $\color{#3D99F6}{t=2^{\cos x}}$ .

$\large 2\color{#3D99F6}{t}^3-5\color{#3D99F6}{t}^2+4\color{#3D99F6}{t}-1=0$

$\large \implies (\color{#3D99F6}{t}-1)^2(2\color{#3D99F6}{t}-1)=0\implies t=2^0,2^{-1}$

$\large \implies \cos x=\color{#D61F06}{0},\color{#20A900}{-1}$

$\large \implies x=\color{#D61F06}{\dfrac{3\pi}{2},\dfrac{\pi}{2}},\color{#20A900}{\pi}$ $\large\text{Their sum}=3\pi\huge \approx \boxed{9.42}$

$\begin{aligned} 2^{1+3\cos x} - 10 \cdot 2^{-1+2 \cos x} + 2^{2+\cos x} - 1 & = 0 \quad \quad \small \color{#3D99F6}{\text{Let }y = 2^{\cos x}} \\ \implies 2y^3 - 5y^2 +4y - 1& = 0 \\ (y-1)^2(2y - 1) & = 0 \\ \implies y & = \begin{cases} 1 \\ \frac{1}{2} \end{cases} \\ 2^{\cos x} & = \begin{cases} 2^0 \\ 2^{-1} \end{cases} \\ \cos x & = \begin{cases} 0 \\ -1 \end{cases} \\ \implies x & = \begin{cases} \frac{\pi}{2}, \ \frac{3\pi}{2} \\ \pi \end{cases} \end{aligned}$

The sum of $x$ satisfying the equation is $\dfrac{\pi}{2} + \dfrac{3\pi}{2} + \pi = 3 \pi \approx \boxed{9.42}$ .