A present to Mehul Arora for his Birthday

Let us express all the four terms in terms of $\cos (2x)$ .

$\begin{cases} \cos (4x) & & = 2 \cos^2 (2x) - 1 \\ \cos^2 (3x) & = (4\cos^3 x - 3 \cos x)^2 & = \frac{1}{2} (4\cos^3 (2x)-3\cos (2x) + 1) \\ \cos^3 (2x) & & = \cos^3 (2x) \\ \cos^4 (x) & = \frac{1}{4}(1+\cos (2x))^2 & = \frac{1}{4}(\cos^2 (2x) + 2\cos (2x)+1) \end{cases}$

Therefore, we have:

$\begin{aligned} \cos(4x)+\cos^2(3x)+\cos^3(2x)+\cos^4(x) & = 0 \\ 3\cos^3(2x) + \frac{9}{4}\cos^2(2x) - \cos (2x) - \frac{1}{4} & = 0 \\ 12\cos^3(2x) + 9\cos^2(2x) - 4\cos (2x) - 1 & = 0 \quad \quad \small \color{#3D99F6}{\text{As }x=\frac{\pi}{2} \text{ is a root,}} \\ (\cos(2x) +1)(12\cos^2 (2x) - 3\cos (2x) -1) & = 0 \end{aligned}$

$\begin{cases} \cos (2x) = - 1 & \Rightarrow 2x = \pm \pi \quad \quad \quad \quad \quad \quad \quad \Rightarrow x = \pm \frac{\pi}{2} & 2 \text{ solutions} \\ \cos (2x) = \frac{3 - \sqrt{57}}{24} & \Rightarrow 2x = \begin{cases} \pm 1.7615 & \Rightarrow x = \pm 0.8808 \\ \pm(2\pi - 1.7615) & \Rightarrow x = \pm 2.2608 \end{cases} & 4 \text{ solutions} \\ \cos (2x) = \frac{3 + \sqrt{57}}{24} & \Rightarrow 2x = \begin{cases} \pm 1.1157 & \Rightarrow x = \pm 0.5578 \\ \pm(2\pi - 1.1157) & \Rightarrow x = \pm 2.5838 \end{cases} & 4 \text{ solutions} \end{cases}$

Therefore, there are $\boxed{10}$ solutions.

P/S: Wondering which Mehul is that.