Ta có $\sin x=\cos 2x\iff \sin x=1-2{\sin }^{2}x\iff \left[\begin{array}{l}\sin x=\frac{1}{2}\\ \sin x=-1\end{array}\right.$.

$+)\sin x=\frac{1}{2}\iff \left[\begin{array}{l}x=\frac{\text{π}}{6}+k2\text{π}\\ x=\frac{5\text{π}}{6}+k2\text{π}\end{array}(k\in \mathbb{Z})\right.$.

+) $\sin x=-1\iff x=-\frac{\text{π}}{2}+\text{k}2\text{π}(\text{k}\in \mathbb{Z})$

Xét $x\in [0;20\text{π}]$ :

Với $x=\frac{\text{π}}{6}+\text{k}2\text{π}$, ta có $0\le \frac{\text{π}}{6}+\text{k}2\text{π}\le 20\text{π}\iff -\frac{1}{12}\le \text{k}\le \frac{119}{12}\Rightarrow$ có 10 giá trị nguyên k thoả mãn.

Với $x=\frac{5\text{π}}{6}+k2\text{π}$, ta có $0\le \frac{5\text{π}}{6}+k2\text{π}\le 20\text{π}\iff -\frac{5}{12}\le \text{k}\le \frac{115}{12}\Rightarrow$ có 10 giá trị nguyên k thoả mãn.

Với $x=-\frac{\text{π}}{2}+\text{k}2\text{π}$, ta có $0\le -\frac{\text{π}}{2}+\text{k}2\text{π}\le 20\text{π}\iff \frac{1}{4}\le \text{k}\le \frac{41}{4}\Rightarrow$ có 10 giá trị nguyên k thoả mãn.