Giải phương trình \(\sin \left( {\pi - x} \right) - \cos \left( {\frac{\pi }{2} - 2x} \right) = 0\)

15/25

\(S = \left\{ {k2\pi |k \in \mathbb{Z}} \right\}\)

\(S = \left\{ {k2\pi ;\frac{\pi }{3} + \frac{{k2\pi }}{3}|k \in \mathbb{Z}} \right\}\).

\(S = \left\{ {k\pi ;\frac{\pi }{3} + \frac{{k2\pi }}{3}|k \in \mathbb{Z}} \right\}\).

\(S = \left\{ {\frac{\pi }{3} + \frac{{k2\pi }}{3}|k \in \mathbb{Z}} \right\}\).

Giải thích

Đáp án đúng là: B

\(\sin \left( {\pi - x} \right) - \cos \left( {\frac{\pi }{2} - 2x} \right) = 0 \Leftrightarrow \sin x - \sin 2x = 0 \Leftrightarrow \sin 2x = \sin x\)

\( \Leftrightarrow \left[ \begin{array}{l}2x = x + k2\pi \\2x = \pi - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \frac{\pi }{3} + \frac{{k2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).