Đề thi giữa kì 1 Toán 11 năm 2023-2024 THPT Nguyễn Thị Minh Khai (TP.HCM) có đáp án

Giải các phương trình sau: a) sin 2x = sin pi/5 ;

Giải thích

a) \[\sin 2x = \sin \frac{\pi }{5}\]

\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{5} + k2\pi \\2x = \pi - \frac{\pi }{5} + k2\pi \end{array} \right.\, \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + k\pi \\x = \frac{{2\pi }}{5} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

b) \[\cot \left( {x + 50^\circ } \right) = \sqrt 3 \]

\( \Leftrightarrow \cot \left( {x + 50^\circ } \right) = \cot 30^\circ \)

\( \Leftrightarrow x + 50^\circ = 30^\circ + k180^\circ \)

\( \Leftrightarrow x = - 20^\circ + k180^\circ \,\,\left( {k \in \mathbb{Z}} \right)\).

c) \(\cos \left( {3x - \frac{\pi }{4}} \right) = \cos 2x\)

\( \Leftrightarrow \left[ \begin{array}{l}3x - \frac{\pi }{4} = 2x + k2\pi \\3x - \frac{\pi }{4} = - 2x + k2\pi \end{array} \right.\, \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{{20}} + k\frac{{2\pi }}{5}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

d) \(\sin \left( {x + 6} \right) = 2{\cos ^2}5x - 1\)

\( \Leftrightarrow \sin \left( {x + 6} \right) = \cos 10x\)

\( \Leftrightarrow \sin \left( {x + 6} \right) = \sin \left( {\frac{\pi }{2} - 10x} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x + 6 = \frac{\pi }{2} - 10x + k2\pi \\x + 6 = \pi - \frac{\pi }{2} + 10x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{22}} - \frac{6}{{11}} + k\frac{{2\pi }}{{11}}\\x = - \frac{\pi }{{18}} + \frac{2}{3} - k\frac{{2\pi }}{9}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).