Find Number of Distinct Solutions to Trigonometric Equation

To find the total number of distinct solutions for the given equation in the domain interval $[0, 2\pi]$:

$\frac{5}{4} \cos^2 2x + \cos^4 x + \sin^4 x + \cos^6 x + \sin^6 x = 2$

We can express the terms involving powers of $\sin x$ and $\cos x$ in terms of $\sin 2x$:

1. Expressing $\cos^4 x + \sin^4 x$: $\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2\left(\frac{\sin 2x}{2}\right)^2 = 1 - \frac{1}{2}\sin^2 2x$

2. Expressing $\cos^6 x + \sin^6 x$: $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x) = 1 - 3 \sin^2 x \cos^2 x = 1 - \frac{3}{4}\sin^2 2x$

Substituting these expressions back into the original equation gives: $\frac{5}{4} \cos^2 2x + \left(1 - \frac{1}{2}\sin^2 2x\right) + \left(1 - \frac{3}{4}\sin^2 2x\right) = 2$

Simplifying the left-hand side: $\frac{5}{4} \cos^2 2x + 2 - \left(\frac{1}{2} + \frac{3}{4}\right)\sin^2 2x = 2$ $\frac{5}{4} \cos^2 2x + 2 - \frac{5}{4}\sin^2 2x = 2$ $\frac{5}{4} \left(\cos^2 2x - \sin^2 2x\right) = 0$

Using the double-angle identity $\cos 4x = \cos^2 2x - \sin^2 2x$: $\frac{5}{4} \cos 4x = 0 \implies \cos 4x = 0$

Given that $x \in [0, 2\pi]$, it follows that $4x \in [0, 8\pi]$.

The general solution for $\cos 4x = 0$ is: $4x = (2k+1)\frac{\pi}{2}, \quad k \in \mathbb{Z}$

For $4x \in [0, 8\pi]$, the possible values for $4x$ are: $4x \in \left\{ \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}, \frac{13\pi}{2}, \frac{15\pi}{2} \right\}$

Dividing by $4$, we get the values of $x$: $x \in \left\{ \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8} \right\}$

All $8$ of these solutions are distinct and fall within the given interval $[0, 2\pi]$.

Thus, the total number of distinct solutions is $8$.