= Solution
Write $s=\widehat{\mathbf n}\cdot\mathbf x$. Taking the <dot product> of the equation with the <unit vector> $\widehat{\mathbf n}$ gives $2s+s^2=-1$, so $(s+1)^2=0$ and $s=-1$.
Now take the <cross product> with $\widehat{\mathbf n}$. The <vector triple product> gives $\widehat{\mathbf n}\times(\widehat{\mathbf n}\times\mathbf x)=s\widehat{\mathbf n}-\mathbf x=-\widehat{\mathbf n}-\mathbf x$, so
$$
2\widehat{\mathbf n}\times\mathbf x-\widehat{\mathbf n}-\mathbf x=\widehat{\mathbf n}\times\mathbf b.
$$
The original equation, now with $s^2=1$, also gives $\widehat{\mathbf n}\times\mathbf x=\mathbf b-\widehat{\mathbf n}-2\mathbf x$. Substitution eliminates the remaining <cross product> of the unknown and yields
$$
\boxed{\mathbf x=\frac{2\mathbf b-3\widehat{\mathbf n}-\widehat{\mathbf n}\times\mathbf b}{5}.}
$$
Its <dot product> with $\widehat{\mathbf n}$ is $-1$, and direct substitution using the <vector triple product> verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the <inverse of a scalar plus a cross-product operator> after the scalar component has been fixed.
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