Grouping: \((\cos x + \cos 4x) + (\cos 2x + \cos 3x) = 0\). Using sum-to-product: \(2\cos\frac{5x}{2}\cos\frac{3x}{2} + 2\cos\frac{5x}{2}\cos\frac{x}{2} = 0\), which gives \(2\cos\frac{5x}{2}(\cos\frac{3x}{2} + \cos\frac{x}{2}) = 0\). Either \(\cos\frac{5x}{2} = 0\) or \(\cos\frac{3x}{2} + \cos\frac{x}{2} = 0\). For \(\cos\frac{5x}{2} = 0\): \(\frac{5x}{2} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}\), giving \(x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}\) (5 solutions). For \(\cos\frac{3x}{2} + \cos\frac{x}{2} = 0\): \(2\cos x \cos\frac{x}{2} = 0\). This gives \(\cos x = 0\) (solutions \(x = \frac{\pi}{2}, \frac{3\pi}{2}\)) or \(\cos\frac{x}{2} = 0\) (solution \(x = \pi\), already counted). Checking: \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) are not in the first set. Total = 5 + 2 = 7.