Using sum-to-product formulas: \((\cos x + \cos 4x) + (\cos 2x + \cos 3x) = 0\). This gives \(2\cos\frac{5x}{2}\cos\frac{3x}{2} + 2\cos\frac{5x}{2}\cos\frac{x}{2} = 0\), which factors as \(2\cos\frac{5x}{2}(\cos\frac{3x}{2} + \cos\frac{x}{2}) = 0\). Either \(\cos\frac{5x}{2} = 0\) or \(\cos\frac{3x}{2} + \cos\frac{x}{2} = 0\). For the first: \(\frac{5x}{2} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}\), giving \(x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}\) (5 solutions). For the second: \(2\cos x \cos\frac{x}{2} = 0\), giving \(\cos x = 0\) or \(\cos\frac{x}{2} = 0\). From \(\cos x = 0\): \(x = \frac{\pi}{2}, \frac{3\pi}{2}\) (2 more solutions not in first set). From \(\cos\frac{x}{2} = 0\): \(x = \pi\) (already counted). Total = 5 + 2 = 7 solutions.