Giải phương trình \(\sin 2x = \sin \left( {x + \frac{\pi }{4}} \right)\)
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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a)
\(\sin \left( {2x + \frac{\pi }{4}} \right) = \sin x \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{4} = x + k2\pi \\2x + \frac{\pi }{4} = \pi - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\3x = \pi - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{4} + \frac{{k2\pi }}{3}\end{array} \right.;k \in Z\)
b)
\(\begin{array}{l}\sin 2x = \cos 3x\\ \Leftrightarrow \cos 3x = \cos \left( {\frac{\pi }{2} - 2x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} - 2x + k2\pi \\3x = - \left( {\frac{\pi }{2} - 2x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\end{array}\)
c)
\(\begin{array}{l}{\cos ^2}2x = {\cos ^2}\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = - \cos \left( {x + \frac{\pi }{6}} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\pi - \left( {x + \frac{\pi }{6}} \right)} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right)\end{array} \right.\end{array}\)
Với \(\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}2x = - \left( {x + \frac{\pi }{6}} \right) + k2\pi \\2x = x + \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{18}} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{6} + k2\pi \end{array} \right.\)
Với \(\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right) \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{6} - x + k2\pi \\2x = - \left( {\frac{{5\pi }}{6} - x} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = \frac{{5\pi }}{6} + k2\pi \\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}\\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right.\)
7 tháng 4 2016
\(\Leftrightarrow\frac{\cos^2x-4\sin^2x.\cos^2x}{4\cos^2x}=\frac{1}{2}\left(\cos\frac{\pi}{3}-\cos2x\right)\)
\(\Leftrightarrow1-4\sin^2x=2\left(\frac{1}{2}-\cos2x\right)\)
\(\Leftrightarrow1-4\sin^2x=1-2\cos2x\)
\(\Leftrightarrow2\sin^2x=\cos2x\)
\(\Leftrightarrow1-\cos2x=\cos2x\)
\(\Leftrightarrow\cos2x=\frac{1}{2}\Leftrightarrow x=\pm\frac{\pi}{6}+k\pi,k\in Z\) thỏa mãn điều kiện
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
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)
\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
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)
\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
d) \(2\cos 3x + 5 = 3\)
\(\begin{array}{l} \Leftrightarrow \cos 3x = - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\cos \left( {3x - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} = - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi + k2\pi }\\{3x = - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)
\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)
b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} = - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi + k2\pi }\\{x = - \pi + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)
4 tháng 10 2020
bài 1 bung công thức sin^6(x) + cos^6(x) là 5/8 + 3/8cos4x = cos4x chuyển vế giải
bài 2 dùng công thức hạ bậc sau đó dùng công thức cộng là ra
ID
11 tháng 5 2022
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=x-\dfrac{\pi}{3}+k2\pi\\3x+\dfrac{\pi}{4}=\pi-\left(x-\dfrac{\pi}{3}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-7\pi}{12}+k2\pi\\4x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7\pi}{24}+k\pi\\x=\dfrac{13\pi}{48}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
11 tháng 5 2022
\(sin\left(3x+\dfrac{\Pi}{4}\right)=sin\left(x-\dfrac{\Pi}{3}\right)\)
\(\Leftrightarrow3x+\dfrac{\Pi}{4}=x-\dfrac{\Pi}{3}+K2\Pi\)
\(\Leftrightarrow2x=-\dfrac{7\Pi}{12}+K2\Pi\)
\(\Leftrightarrow x=-\dfrac{7\Pi}{24}+K\Pi\) \(\left(K\in Z\right)\)
\(\sin 2x = \sin \left( {x + \frac{\pi }{4}} \right) \Leftrightarrow \left[ \begin{array}{l}2x = x + \frac{\pi }{4} + k2\pi \\2x = \pi - \left( {x + \frac{\pi }{4}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\3x = \frac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{4} + \frac{{k2\pi }}{3}\end{array} \right.\)