Worked solution: Wantzel's algebraic impossibility proof via field extensions (1837)

Trisecting an arbitrary angle $\theta$ with straightedge and compass means constructing an angle of $\theta/3$ starting from $\theta$, using only the two classical tools. Wantzel's 1837 paper, published in Liouville's Journal de mathématiques pures et appliquées, settled this question alongside doubling the cube by reducing both to algebra: a real number is constructible exactly when it can be reached from $\mathbb{Q}$ by a finite chain of square-root extractions (Wantzel 1837, §I).

The plan carried out in the remaining steps is: (1) show every constructible length has degree a power of $2$ over $\mathbb{Q}$; (2) reduce trisecting $60^\circ$ to solving a cubic equation for $x=2\cos 20^\circ$; (3) show that cubic is irreducible over $\mathbb{Q}$, so its root has degree $3$; (4) since $3$ is not a power of $2$, $\cos 20^\circ$ is not constructible, so $60^\circ$ cannot be trisected.

Because one counterexample suffices to kill a general method, this single angle already proves that no straightedge-and-compass procedure can trisect every angle, even though some special angles (like $90^\circ$) can be trisected.