Giải các phương trình sau:
a) \(\cos 2x = 1;\)
b) \(\sin \left( {x - \frac{\pi }{4}} \right) = - 1;\)
c) \(\cos \left( {4x - {{75}^0}} \right) = - \frac{{\sqrt 3 }}{2};\)
d) \(\sin \left( {3x - {{15}^0}} \right) = 0.\)
a) \(\cos a = 1 \Leftrightarrow a = k2\pi \left( {k \in \mathbb{Z}} \right)\)
b) \(\sin a = - 1 \Leftrightarrow a = - \frac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right)\)
c)
\(\begin{array}{l}\cos x = m\\ \Leftrightarrow \cos x = \cos \alpha \\ \Rightarrow \left[ \begin{array}{l}x = \alpha + k2\pi \\x = - \alpha + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
d) \(\sin a = 0 \Leftrightarrow a = k\pi \left( {k \in \mathbb{Z}} \right)\)
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a)
\(\begin{array}{l}\cos 2x = 1\\ \Leftrightarrow 2x = k2\pi \left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow x = k\pi \left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là \(x = k\pi \left( {k \in \mathbb{Z}} \right)\)
b)
\(\begin{array}{l}\sin \left( {x - \frac{\pi }{4}} \right) = - 1\\ \Leftrightarrow x - \frac{\pi }{4} = - \frac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow x = - \frac{\pi }{4} + k2\pi \left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là \(x = - \frac{\pi }{4} + k2\pi \left( {k \in \mathbb{Z}} \right)\)
c)
\(\begin{array}{l}\cos \left( {4x - {{75}^0}} \right) = - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \cos \left( {4x - {{75}^0}} \right) = \cos 150{}^0\\ \Leftrightarrow \left[ \begin{array}{l}4x - {75^0} = {150^0} + k{360^0}\\4x - {75^0} = - {150^0} + k{360^0}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}4x = {225^0} + k{360^0}\\4x = - {75^0} + k{360^0}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = {\left( {\frac{{225}}{4}} \right)^0} + k{90^0}\\x = {\left( { - \frac{{75}}{4}} \right)^0} + k{90^0}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là \(x = {\left( {\frac{{225}}{4}} \right)^0} + k{90^0},x = {\left( { - \frac{{75}}{4}} \right)^0} + k{90^0}\left( {k \in \mathbb{Z}} \right)\)
d)
\(\begin{array}{l}\sin \left( {3x - {{15}^0}} \right) = 0\\ \Leftrightarrow 3x - {15^0} = k{180^0}\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow 3x = {15^0} + k{180^0}\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow x = {5^0} + k{60^0}\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là \(x = {5^0} + k{60^0}\left( {k \in \mathbb{Z}} \right)\)