Bài 3 trang 28 SGK Đại số và Giải tích 11

Đề bài

Giải các phương trình sau:

\(\begin{array}{l}
a)\,\,\cos \left( {x - 1} \right) = \frac{2}{3}\\
b)\,\,\cos 3x = \cos {12^0}\\
c)\,\,\cos \left( {\frac{{3x}}{2} - \frac{\pi }{4}} \right) = - \frac{1}{2}\\
d)\,\,{\cos ^2}2x = \frac{1}{4}
\end{array}\)

Hướng dẫn giải

\(\cos x = \cos \alpha \Leftrightarrow \left[ \begin{array}{l}
x = \alpha + k2\pi \\
x = - \alpha + k2\pi
\end{array} \right.\,\,\left( {k \in Z} \right)\)

Lời giải chi tiết

\(\begin{array}{l}
a)\,\,\cos \left( {x - 1} \right) = \frac{2}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = \arccos \frac{2}{3} + k2\pi \\
x - 1 = - \arccos \frac{2}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arccos \frac{2}{3} + 1 + k2\pi \\
x = - \arccos \frac{2}{3} + 1 + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
b)\,\,\cos 3x = \cos {12^0}\\
\Leftrightarrow \left[ \begin{array}{l}
3x = {12^0} + k{360^0}\\
3x = - {12^0} + k{360^0}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = {4^0} + k{120^0}\\
x = - {4^0} + k{120^0}
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
c)\,\,\cos \left( {\frac{{3x}}{2} - \frac{\pi }{4}} \right) = - \frac{1}{2}\\
\Leftrightarrow \cos \left( {\frac{{3x}}{2} - \frac{\pi }{4}} \right) = \cos \frac{{2\pi }}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{{3x}}{2} - \frac{\pi }{4} = \frac{{2\pi }}{3} + k2\pi \\
\frac{{3x}}{2} - \frac{\pi }{4} = - \frac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{{3x}}{2} = \frac{{11\pi }}{{12}} + k2\pi \\
\frac{{3x}}{2} = - \frac{{5\pi }}{{12}} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{11\pi }}{{18}} + \frac{{4k\pi }}{3}\\
x = \frac{{ - 5\pi }}{{18}} + \frac{{4k\pi }}{3}
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
d)\,\,{\cos ^2}2x = \frac{1}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \frac{1}{2} = \cos \frac{\pi }{3}\\
\cos 2x = - \frac{1}{2} = \cos \frac{{2\pi }}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \pm \frac{\pi }{3} + k2\pi \\
2x = \pm \frac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \frac{\pi }{6} + k\pi \\
x = \pm \frac{\pi }{3} + k\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)