Giải bài 3 trang 28 SGK Đại số và Giải tích 11

LG a

\(\begin{array}{l}\,\,\cos \left( {x - 1} \right) = \frac{2}{3}\\\end{array}\)

Phương pháp 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:

\(\begin{array}{l}
\,\,\cos \left( {x - 1} \right) = \frac{2}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = \arccos \frac{2}{3} + k2\pi \\
x - 1 = - \arccos \frac{2}{3} + k2\pi 
\end{array} \right.( {k \in \mathbb{Z}})\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arccos \frac{2}{3} + 1 + k2\pi \\
x = - \arccos \frac{2}{3} + 1 + k2\pi 
\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\\end{array}\)

LG b

\(\begin{array}{l} \,\,\cos 3x = \cos {12^0}\\\end{array}\)

Phương pháp giải:

\(\cos x = \cos a \Leftrightarrow \left[ \begin{array}{l}
x = a + k360^0 \\
x = - a + k360^0 
\end{array} \right.\,\,\left( {k \in Z} \right)\)

Lời giải chi tiết:

\(\begin{array}{l}\,\,\cos 3x = \cos {12^0}\\
\Leftrightarrow \left[ \begin{array}{l}
3x = {12^0} + k{360^0}\\
3x = - {12^0} + k{360^0}
\end{array} \right.( {k \in \mathbb{Z}})\\
\Leftrightarrow \left[ \begin{array}{l}
x = {4^0} + k{120^0}\\
x = - {4^0} + k{120^0}
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\\end{array}\)

LG c

\(\begin{array}{l} \,\,\cos \left( {\frac{{3x}}{2} - \frac{\pi }{4}} \right) = - \frac{1}{2}\\\end{array}\)

Phương pháp 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:

\(\begin{array}{l}\,\,\cos \left( {\frac{{3x}}{2} - \frac{\pi }{4}} \right) = - \frac{1}{2}\\
\Leftrightarrow \cos \left( {\frac{{3x}}{2} - \frac{\pi }{4}} \right) = \cos \frac{{2\pi }}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{{3x}}{2} - \frac{\pi }{4} = \frac{{2\pi }}{3} + k2\pi \\
\frac{{3x}}{2} - \frac{\pi }{4} = - \frac{{2\pi }}{3} + k2\pi 
\end{array} \right.( {k \in \mathbb{Z}})\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{{3x}}{2} = \frac{{11\pi }}{{12}} + k2\pi \\
\frac{{3x}}{2} = - \frac{{5\pi }}{{12}} + k2\pi 
\end{array} \right.( {k \in \mathbb{Z}})\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{11\pi }}{{18}} + \frac{{4k\pi }}{3}\\
x = \frac{{ - 5\pi }}{{18}} + \frac{{4k\pi }}{3}
\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\\end{array}\)

LG d

\(\begin{array}{l} \,\,{\cos ^2}2x = \frac{1}{4}
\end{array}\)

Phương pháp 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:

\(\begin{array}{l}\,\,{\cos ^2}2x = \frac{1}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \frac{1}{2} = \cos \frac{\pi }{3}\\
\cos 2x = - \frac{1}{2} = \cos \frac{{2\pi }}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \pm \frac{\pi }{3} + k2\pi \\
2x = \pm \frac{{2\pi }}{3} + k2\pi 
\end{array} \right.( {k \in \mathbb{Z}})\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \frac{\pi }{6} + k\pi \\
x = \pm \frac{\pi }{3} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)

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