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

$$ \begin{aligned} \Rightarrow & \frac{5}{4} \cos ^2 2 x+\left(\cos ^2 x\right)^2+\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^3 \\\\ & +\left(\sin ^2 x\right)^3=2\quad\quad...(i) \end{aligned} $$

As we know, $a^2+b^2+2 a b=(a+b)^2$

$$ \Rightarrow a^2+b^2=(a+b)^2-2 a b $$

And $a^3+b^3+3 a b(a+b)=(a+b)^3$

$$ \Rightarrow a^3+b^3=(a+b)^3-3 a b(a+b) $$

So, equation (i) can be written as

$$ \begin{aligned} & \frac{5}{4} \cos ^2 2 x+\left(\cos ^2 x+\sin ^2 x\right)^2-2\left(\cos ^2 x\right) \\\\ &\left(\sin ^2 x\right)+\left(\cos ^2 x+\sin ^2 x\right)^3-3 \cos ^2 x \sin ^2 x \\\\ &\left(\cos ^2 x+\sin ^2 x\right)=2 \\\\ & \Rightarrow \frac{5}{4} \cos ^2 2 x+(1)^2-2 \cos ^2 x \sin ^2 x+(1)^3 \\\\ &-3 \cos ^2 x \sin ^2 x(1)=2~~~~\left\{\because \cos ^2 x+\sin ^2 x=1\right\} \end{aligned} $$

$$ \begin{aligned} & \Rightarrow \frac{5}{4} \cos ^2 2 x+2-5 \cos ^2 x \sin ^2 x=2 \\\\ & \Rightarrow \frac{5}{4} \cos ^2 2 x-5 \cos ^2 x \sin ^2 x=0 \end{aligned} $$

As we know, $\sin 2 \theta=2 \sin \theta \cos \theta$

$$ \begin{aligned} & \Rightarrow \frac{5}{4} \cos ^2 2 x-\frac{5}{4} \sin ^2 2 x=0 \\\\ & \because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta \\\\ & \Rightarrow \frac{5}{4} \cos 4 x=0 \\\\ & \Rightarrow \cos 4 x=0 \\\\ & \Rightarrow 4 x=2(n+1) \frac{\pi}{2}, n \in \mathrm{I} \\\\ & \because x \in[0,2 \pi] \end{aligned} $$

So, possible distinct values of $x$ are $\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}$ and $\frac{15 \pi}{8}$.

So, the number of distinct solutions of the given equation are 8 .