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