Given equation

We need to solve $\frac{5}{4}\cos^2 2x + \cos^4 x + \sin^4 x + \cos^6 x + \sin^6 x = 2$ for $x \in [0,2\pi]$.

Simplify $\cos^4 x + \sin^4 x$

Using $a^2+b^2=(a+b)^2-2ab$ with $a=\cos^2 x$, $b=\sin^2 x$, $\cos^4 x+\sin^4 x=(\cos^2 x+\sin^2 x)^2-2\sin^2 x\cos^2 x$ $=1-2\sin^2 x\cos^2 x$ Since $\sin^2 2x=4\sin^2 x\cos^2 x,$ we get $\cos^4 x+\sin^4 x=1-\frac{1}{2}\sin^2 2x$ Also, because $\sin^2 2x=1-\cos^2 2x$, $\cos^4 x+\sin^4 x=1-\frac{1}{2}(1-\cos^2 2x)=\frac{1}{2}+\frac{1}{2}\cos^2 2x.$

Simplify $\cos^6 x + \sin^6 x$

Use $a^3+b^3=(a+b)^3-3ab(a+b)$ with $a=\cos^2 x$, $b=\sin^2 x$: $\cos^6 x+\sin^6 x=(\cos^2 x+\sin^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)$ $=1-3\sin^2 x\cos^2 x$ $=1-\frac{3}{4}\sin^2 2x$ $=1-\frac{3}{4}(1-\cos^2 2x)=\frac{1}{4}+\frac{3}{4}\cos^2 2x.$

Substitute into the equation

The left-hand side becomes $\frac{5}{4}\cos^2 2x + \left(\frac{1}{2}+\frac{1}{2}\cos^2 2x\right) + \left(\frac{1}{4}+\frac{3}{4}\cos^2 2x\right).$

Combine constants and coefficients of $\cos^2 2x$: $\left(\frac{1}{2}+\frac{1}{4}\right) + \left(\frac{5}{4}+\frac{1}{2}+\frac{3}{4}\right)\cos^2 2x = 2$ $\frac{3}{4} + \frac{10}{4}\cos^2 2x = 2$ $\frac{3}{4} + \frac{5}{2}\cos^2 2x = 2.$

Solve for $\cos^2 2x$

$\frac{5}{2}\cos^2 2x = 2-\frac{3}{4} = \frac{5}{4}$ $\cos^2 2x = \frac{5}{4}\cdot \frac{2}{5} = \frac{1}{2}.$

So, $\cos 2x = \pm \frac{1}{\sqrt{2}}.$

Solve in $[0,2\pi]$

We need $2x = \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4} \pmod{2\pi}.$

Since $x \in [0,2\pi]$, we have $2x \in [0,4\pi]$. Therefore all solutions for $2x$ in $[0,4\pi]$ are: $\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4},\frac{9\pi}{4},\frac{11\pi}{4},\frac{13\pi}{4},\frac{15\pi}{4}.$

Dividing by $2$: $x=\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8},\frac{13\pi}{8},\frac{15\pi}{8}.$

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

Comparison with stored answer

Derived answer = $8$.

Stored correct answer = $8$.

Hence, they agree.