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Ta có:
\(\begin{array}{l}cos2x = cos\left( {x + \frac{\pi }{3}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = x + \frac{\pi }{3} + k2\pi \\2x = - x - \frac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
Với \(x = \frac{\pi }{3} + k2\pi \),\(k \in \mathbb{Z}\) đạt giá trị âm lớn nhất khi k = – 1, khi đó \(x = \frac{\pi }{3} - 2\pi = \frac{{ - 5\pi }}{3}\)
Với \(x = - \frac{\pi }{9} + k\frac{{2\pi }}{3}\),\(k \in \mathbb{Z}\) đạt giá trị âm lớn nhất khi k = 0, khi đó \(x = x = - \frac{\pi }{9} + 0.\frac{{2\pi }}{3} = - \frac{\pi }{9}\)
Vậy nghiệm âm lớn nhất của phương trình đã cho là \( - \frac{\pi }{9}\).
Đáp án: A
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\begin{array}{l}a)\;\,cos(x + \frac{\pi }{3}) = \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow cos\left( {x + \frac{\pi }{3}} \right) = cos\frac{\pi }{6}\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{3} = \frac{\pi }{6} + k2\pi \\x + \frac{\pi }{3} = -\frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = -\frac{\pi }{6} + k2\pi \\x = -\frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
\(\begin{array}{l}b)\;\,cos4x = cos\frac{{5\pi }}{{12}}\\ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{{5\pi }}{{12}} + k2\pi \\4x = -\frac{{5\pi }}{{12}} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{48}} + k\frac{\pi }{2}\\x = -\frac{{5\pi }}{{48}} + k\frac{\pi }{2}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
\(\begin{array}{l}c)\;\,co{s^2}x = 1\\ \Leftrightarrow \left[ \begin{array}{l}cosx = 1\\cosx = -1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \pi + k2\pi \end{array} \right. \Leftrightarrow x = k\pi ,k \in \mathbb{Z}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)
\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha = \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = \pm \frac{{2\sqrt 2 }}{3}\end{array}\)
Vì \( - \frac{\pi }{2} < \alpha < 0\) nên \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Điều kiện xác định: \(\frac{1}{2}x + \frac{\pi }{4} \ne k\pi \Leftrightarrow x \ne - \frac{\pi }{2} + k2\pi ,k \in \mathbb{Z}.\)
Ta có: \(cot\left( {\frac{1}{2}x + \frac{\pi }{4}} \right) = - 1 \Leftrightarrow cot\left( {\frac{1}{2}x + \frac{\pi }{4}} \right) = \cot \left( { - \frac{\pi }{4}} \right)\)
\( \Leftrightarrow \frac{1}{2}x + \frac{\pi }{4} = - \frac{\pi }{4} + k\pi \Leftrightarrow x = - \pi + k2\pi ,k \in \mathbb{Z}\,\,(TM).\)
Vậy \(x = - \pi + k2\pi ,k \in \mathbb{Z}\,\).
b, Điều kiện xác định: \(3x \ne k\pi \Leftrightarrow x \ne k\frac{\pi }{3},k \in \mathbb{Z}.\)
\(\;cot3x = - \frac{{\sqrt 3 }}{3} \Leftrightarrow cot3x = \cot \left( { - \frac{\pi }{3}} \right)\)
\( \Leftrightarrow 3x = - \frac{\pi }{3} + k\pi \Leftrightarrow x = - \frac{\pi }{9} + k\frac{\pi }{3},k \in \mathbb{Z}\,\,(TM).\)
Vậy \(x = - \frac{\pi }{9} + k\frac{\pi }{3},k \in \mathbb{Z}\,\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Với mọi \(x \in \mathbb{R}\) ta có \( - 1 \le cosx \le 1\)
Vậy phương trình \(cosx = - 3\;\) vô nghiệm.
\(\begin{array}{l}b)\,\;cosx = cos{15^o}\;\\ \Leftrightarrow \left[ \begin{array}{l}x = {15^o} + k{360^o},k \in \mathbb{Z}\\x = - {15^o} + k{360^o},k \in \mathbb{Z}\end{array} \right.\end{array}\)
Vậy phương trình có nghiệm \(x = {15^o} + k{360^o}\) hoặc \(x = - {15^o} + k{360^o},k \in \mathbb{Z}\).
\(\begin{array}{l}c)\;\,cos(x + \frac{\pi }{{12}}) = cos\frac{{3\pi }}{{12}}\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{{12}} = \frac{{3\pi }}{{12}} + k2\pi ,k \in \mathbb{Z}\\x + \frac{\pi }{{12}} = - \frac{{3\pi }}{{12}} + k2\pi ,k \in \mathbb{Z}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi ,k \in \mathbb{Z}\\x = - \frac{\pi }{3} + k2\pi ,k \in \mathbb{Z}\end{array} \right.\end{array}\)
Vậy phương trình có nghiệm \(x = \frac{\pi }{6} + k2\pi ,\) hoặc \(x = - \frac{\pi }{3} + k2\pi ,k \in \mathbb{Z}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Điều kiện xác định: \(x \ne 90^\circ + k180^\circ \).
Ta có:\({\rm{ }}tanx = tan55^\circ \Leftrightarrow x = 55^\circ + k180^\circ ,{\rm{ }}k\; \in \;\mathbb{Z}\,\,(TM).\)
b, Điều kiện xác định: \(2x + \frac{\pi }{4} \ne \frac{\pi }{2} + k\pi \Leftrightarrow x \ne \frac{\pi }{8} + k\pi ,k \in \mathbb{Z}.\)
Ta có: \(\tan \left( {2x + \frac{\pi }{4}} \right) = 0 \Leftrightarrow 2x + \frac{\pi }{4} = k\pi \Leftrightarrow x = -\frac{\pi }{8} + k\frac{\pi }{2},k \in \mathbb{Z}\,\,(TM).\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
+) Xét \(\beta = - \alpha \), khi đó:
\(\begin{array}{l}cos\beta = cos\left( {-{\rm{ }}\alpha } \right) = cos\alpha ;\\sin\beta = sin\left( {-{\rm{ }}\alpha } \right) = -sin\alpha \Leftrightarrow sin\alpha = -sin\beta .\end{array}\)
Do đó A thỏa mãn.
Đáp án: A
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
b) \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c)
\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)
d)
\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x = - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)
e)
\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)
f)
\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)
a) Vì \(\sin \frac{\pi }{6} = \frac{1}{2}\) nên ta có phương trình \(sin2x = \sin \frac{\pi }{6}\)
\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{6} + k2\pi \\2x = \pi - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
\(\begin{array}{l}b,\,\,sin(x - \frac{\pi }{7}) = sin\frac{{2\pi }}{7}\\ \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{7} = \frac{{2\pi }}{7} + k2\pi \\x - \frac{\pi }{7} = \pi - \frac{{2\pi }}{7} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{3\pi }}{7} + k2\pi \\x = \frac{{6\pi }}{7} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
\(\begin{array}{l}\;c)\;sin4x - cos\left( {x + \frac{\pi }{6}} \right) = 0\\ \Leftrightarrow sin4x = cos\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{2} - x - \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{3} - x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{3} - x + k2\pi \\4x = \pi - \frac{\pi }{3} + x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{15}} + k\frac{{2\pi }}{5}\\x = \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)