Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
{\cos ^{2019}}x + {\sin ^{2020}}x = 1\\
\Leftrightarrow {\cos ^{2019}}x + {\sin ^{2020}}x = {\cos ^2}x + {\sin ^2}x\\
\Leftrightarrow {\cos ^2}x({\cos ^{2017}}x - 1) = {\sin ^2}x(1 - {\sin ^{2018}}x)(*)\\
- 1 \le \cos x \le 1 \Leftrightarrow - 1 \le {\cos ^{2017}}x \le 1 \Rightarrow {\cos ^{2017}}x - 1 \le 0;{\cos ^2}x \ge 0 \Rightarrow VT \le 0\\
- 1 \le \sin x \le 1 \Leftrightarrow 0 \le {\sin ^{2018}}x \le 1 \Rightarrow 1 - {\sin ^{2018}}x \ge 0;{\sin ^2}x \ge 0 \Rightarrow VP \ge 0\\
\Rightarrow (*) \Leftrightarrow \left\{ \begin{array}{l}
VP = 0\\
VT = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\sin ^2}x(1 - {\sin ^{2018}}x) = 0\\
{\cos ^2}x({\cos ^{2017}}x - 1) = 0
\end{array} \right. \Leftrightarrow x = \frac{{k\pi }}{2}
\end{array}\]
