To solve this problem, we are given that:

• \(\cos\alpha + \cos\beta = 0\)
• \(\sin\alpha + \sin\beta = 0\)

We need to find the value of \(\cos 2\alpha + \cos 2\beta + 2 \cos(\alpha + \beta)\).

From the given equations, we can deduce the following trigonometric identities. If \(\cos\alpha + \cos\beta = 0\), then:

\(\cos\alpha = -\cos\beta\)

Similarly, from \(\sin\alpha + \sin\beta = 0\), it follows that:

\(\sin\alpha = -\sin\beta\)

By using the sum-to-product identities for trigonometric functions, the following holds true:

\(\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\)

Since \(\cos\alpha = -\cos\beta\) and \(\sin\alpha = -\sin\beta\), we obtain:

\(\cos(\alpha + \beta) = (-\cos\beta)\cos\beta - (-\sin\beta)\sin\beta\)

\(= -\cos^2\beta + \sin^2\beta\)

\(= -1\) (since \(\cos^2\beta + \sin^2\beta = 1\))

We must find:

\(\cos 2\alpha + \cos 2\beta + 2 \cos(\alpha + \beta)\)

Recall that \(\cos 2\theta = 2\cos^2\theta - 1\), so:

\(\cos 2\alpha = 2\cos^2\alpha - 1\) and \(\cos 2\beta = 2\cos^2\beta - 1\)

Thus:

\(\cos 2\alpha + \cos 2\beta = (2\cos^2\alpha - 1) + (2\cos^2\beta - 1)\)

\(= 2(\cos^2\alpha + \cos^2\beta) - 2\)

Combining the identities, we find:

\(\cos 2\alpha + \cos 2\beta + 2 \cos(\alpha + \beta)\)

\(= (2(\cos^2\alpha + \cos^2\beta) - 2) + 2(-1)\)

\(= 2(\cos^2\alpha + \cos^2\beta) - 4\)

Recall \(\cos^2\alpha + \sin^2\alpha = 1\), and similarly \(\cos^2\beta + \sin^2\beta = 1\), thus:

\(\cos^2\beta = \frac{1}{2}\) and \(\cos^2\alpha = \frac{1}{2}\)

Thus, solving further:

\(2(\frac{1}{2} + \frac{1}{2}) - 4 = 0\)

Final conclusion: The value of \(\cos 2\alpha + \cos 2\beta + 2 \cos(\alpha + \beta)\) is 0.