De acordo com o Teorema de Laplace: $\begin{vmatrix} \log x & \log x & \log x\\ \log 6x & \log 3x & \cos x\\ 1 & 1 & \log^{2} x \end{vmatrix}=(D_{11}\cdot \log x)-(D_{21}\cdot \log 6x)+D_{31}$ $=$ $D$ $=$ $0$ $D_{11}=\log^{2} x \cdot \log 3x - \cos x $ $\implies$ $(D_{11}\cdot \log x)=\log^{3} x \cdot \log 3x - \cos x \log x $ $D_{21}=\log^{3} x - \log x$ $\implies$ $(D_{21}\cdot \log 6x)=\log 6x \log^{3} x - \log 6x \log x$ $D_{31}=\cos x \cdot \log x - \log 3x \cdot \log x$ $D=\log^{3} x \cdot( \log 3x- \log 6x) - \cancel {\cos x \log x} +\log x \cdot ( \log 6x - \log 3x) + \cancel {\cos x \cdot \log x}$ $D=\log^{3} x \cdot( \cancel{\log 3x} - \cancel{\log 3x} - \log 2) +\log x \cdot ( \log 2 + \cancel{\log 3x} - \cancel{\log 3x})$ $D=-\log 2 \cdot \log^{3} x + \log 2 \cdot \log x$ $=$ $0$, isto implica em: $\log 2 \cdot \log^{3} x = \log 2 \cdot \log x$ $(i)$ $\log x = 0$ $\implies$ $x=1$ $(ii)$ $\log x \neq 0$ $\implies$ $\log^{2} x = 1$ $\implies$ $\log x = \pm 1$ $\implies$ $x=10^{\pm 1} = \{\frac{1}{10}, 10\}$ Soluções: $x=\{\frac{1}{10},1, 10\}$; Soma das soluções: $\boxed { \frac{1}{10}+1+10=11,1 }$ Alternativa $(\mathbb{E})$