A geometry problem by Jose Sacramento

$\begin{aligned} 2\cos x + 3 \sin x & = 2 & \small {\color{#3D99F6}\text{Divide both sides by }\cos x} \\ 2 + 3 \tan x & = 2 \sec x & \small {\color{#3D99F6}\text{Square both sides}} \\ 4 + 12 \tan x + 9 \tan^2 x & = 4 \sec^2 x \\ 4 + 12 \tan x + 9 \tan^2 x & = 4 + 4\tan^2 x \\ 12 \tan x + 5 \tan^2 x & = 0 \\ \tan x (5\tan x + 12) & = 0 \end{aligned}$

$\implies \tan x = \begin{cases} 0 & \implies x = 0 \\ -\frac {12}5 & \implies x \approx 1.9656 \end{cases}$

The sum of solutions is therefore $\approx \boxed{1.96}$ .