Bài 11 trang 42 SGK Toán 11 tập 1 – Chân trời sáng tạo

Đề bài

Giải các phương trình sau:

$\begin{array}{l}a)\;sin2x + cos3x = 0\\b)\;sinx.cosx = \frac{{\sqrt 2 }}{4}\\c)\;sinx + sin2x = 0\end{array}$

Lời giải chi tiết

$\begin{array}{l}a)\;sin2x + cos3x = 0\\ \Leftrightarrow cos\left( {\frac{\pi }{2} – 2x} \right) + cos3x = 0\\ \Leftrightarrow cos\left( {\frac{\pi }{2} – 2x} \right) =  – cos3x\\ \Leftrightarrow cos\left( {\frac{\pi }{2} – 2x} \right) = cos\left( {\pi  – 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{\pi }{2} – 2x = \pi  – 3x + k2\pi \\\frac{\pi }{2} – 2x =  – \pi  + 3x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x = \frac{{3\pi }}{{10}} + k\frac{{2\pi }}{5}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}$

$\begin{array}{l}b)\;sinx.cosx = \frac{{\sqrt 2 }}{4}\\ \Leftrightarrow \frac{1}{2}\;sin2x = \frac{{\sqrt 2 }}{4}\\ \Leftrightarrow sin2x = \frac{{\sqrt 2 }}{2} = sin\left( {\frac{\pi }{4}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{4} + k2\pi \\2x = \pi  – \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{8} + k\pi \\x = \frac{{3\pi }}{8} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}$

$\begin{array}{l}c)\;sinx + sin2x = 0\\ \Leftrightarrow sinx =  – sin2x\\ \Leftrightarrow sinx = sin( – 2x)\\ \Leftrightarrow \left[ \begin{array}{l}x =  – 2x + k2\pi \\x = \pi  + 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\frac{{2\pi }}{3}\\x =  – \pi  + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}$