1. We need to find $S=\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right).$

2. Use the identity $1+\omega+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6=0,$ where $\omega=e^{\frac{2\pi i}{7}}.$

Taking real parts of both sides: $1+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{10\pi}{7}+\cos\frac{12\pi}{7}=0.$

3. Now use periodicity and symmetry of cosine: $\cos\frac{8\pi}{7}=\cos\left(2\pi-\frac{6\pi}{7}\right)=\cos\frac{6\pi}{7},$ $\cos\frac{10\pi}{7}=\cos\left(2\pi-\frac{4\pi}{7}\right)=\cos\frac{4\pi}{7},$ $\cos\frac{12\pi}{7}=\cos\left(2\pi-\frac{2\pi}{7}\right)=\cos\frac{2\pi}{7}.$

So the real-part equation becomes $1+2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)=0.$ That is, $1+2S=0.$

4. Hence, $S=-\frac{1}{2}.$

5. Checking options:

• A: $-1$ ❌
• B: $-\frac12$ ✅
• C: $-\frac13$ ❌
• D: $-\frac14$ ❌

Therefore, the correct answer is Option B.