cos(π/7) − cos(2π/7) + cos(3π/7) = 1/2

Rewriting. $-\cos\frac{2\pi}7=\cos\frac{5\pi}7$. So the sum is
$$S=\cos a+\cos3a+\cos5a,\qquad a=\frac\pi7.$$

Multiplying by $2\sin a$. For each $k$, $2\sin a\cos\bigl((2k-1)a\bigr)=\sin(2ka)-\sin\bigl((2k-2)a\bigr)$. Summing for $k=1,2,3$ the right side telescopes:
$$2\sin a\cdot S=\sin6a-\sin0=\sin\frac{6\pi}7=\sin\frac\pi7.$$

Conclusion. Since $\sin\frac\pi7\ne0$, we get $S=\frac12$.

Check. Numerically $0.90097-0.62349+0.22252=0.5000$ ✓.

Lesson. A sum of cosines of an arithmetic progression of angles is tamed by multiplying by $2\sin$ of half the common difference.