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:

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

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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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