5. The equation
$$
\left(\sin ^{4} x+\sin ^{2} x\right)^{3}-\left(\cos ^{4} x+\cos ^{2} x\right)^{3}=2 \cos 2 x
$$

has the solution set $\qquad$ .
5. $\left\{\left.\frac{(2 k+1) \pi}{4} \right\rvert\, k \in \mathbf{Z}\right\}$.

The original equation is
$$
\begin{array}{l}
\left(\sin ^{4} x+\sin ^{2} x\right)^{3}-\left(\cos ^{4} x+\cos ^{2} x\right)^{3} \\
=2\left(\cos ^{2} x-\sin ^{2} x\right) \\
=\left(\sin ^{2} x+\cos ^{2} x+1\right)\left(\sin ^{2} x-\cos ^{2} x\right) \\
\Rightarrow\left(\sin ^{4} x+\sin ^{2} x\right)^{3}+\left(\sin ^{4} x+\sin ^{2} x\right) \\
=\left(\cos ^{4} x+\cos ^{2} x\right)^{3}+\left(\cos ^{4} x+\cos ^{2} x\right) .
\end{array}
$$

Notice that the function $f(x)=x^{3}+x$ is strictly increasing.
$$
\begin{array}{l}
\text { Hence } \sin ^{4} x+\sin ^{2} x=\cos ^{4} x+\cos ^{2} x \\
\Rightarrow \cos 2 x=0 \\
\Rightarrow x \in\left\{\left.\frac{(2 k+1) \pi}{4} \right\rvert\, k \in \mathbf{Z}\right\} .
\end{array}
$$