Toán Lớp 11: giải phương trình lượng giác sau:
a)Sin^22x+cos2x=0
b) cot2x-tan2x+1=0
c) sin2x+2cosx=0
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Toán Lớp 11: giải phương trình lượng giác sau:
a)Sin^22x+cos2x=0
b) cot2x-tan2x+1=0
c) sin2x+2cosx=0
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = \dfrac{1}{2}\arccos \dfrac{{1 – \sqrt 5 }}{2} + k\pi \\
x = – \dfrac{1}{2}\arccos \dfrac{{1 – \sqrt 5 }}{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{1}{2}\arctan \dfrac{{1 – \sqrt 5 }}{2} + k\pi \\
x = \dfrac{1}{2}\arctan \dfrac{{1 + \sqrt 5 }}{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
x = \dfrac{\pi }{2} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
a,\\
{\sin ^2}2x + \cos 2x = 0\\
\Leftrightarrow \left( {1 – {{\cos }^2}2x} \right) + \cos 2x = 0\\
\Leftrightarrow – {\cos ^2}2x + \cos 2x + 1 = 0\\
\Leftrightarrow {\cos ^2}2x – \cos 2x – 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \dfrac{{1 + \sqrt 5 }}{2}\\
\cos 2x = \dfrac{{1 – \sqrt 5 }}{2}
\end{array} \right.\\
– 1 \le \cos 2x \le 1 \Rightarrow \cos 2x = \dfrac{{1 – \sqrt 5 }}{2}\\
\cos 2x = \dfrac{{1 – \sqrt 5 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \arccos \dfrac{{1 – \sqrt 5 }}{2} + k2\pi \\
2x = – \arccos \dfrac{{1 – \sqrt 5 }}{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\arccos \dfrac{{1 – \sqrt 5 }}{2} + k\pi \\
x = – \dfrac{1}{2}\arccos \dfrac{{1 – \sqrt 5 }}{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\sin 2x \ne 0\\
\cos 2x \ne 0
\end{array} \right. \Leftrightarrow 2x \ne \dfrac{{k\pi }}{2} \Leftrightarrow x \ne \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)\\
\cot 2x – \tan 2x + 1 = 0\\
\Leftrightarrow \dfrac{1}{{\tan 2x}} – \tan 2x + 1 = 0\\
\Leftrightarrow \dfrac{{1 – {{\tan }^2}2x + \tan 2x}}{{\tan 2x}} = 0\\
\Leftrightarrow 1 – {\tan ^2}2x + \tan 2x = 0\\
\Leftrightarrow {\tan ^2}2x – \tan 2x – 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan 2x = \dfrac{{1 – \sqrt 5 }}{2}\\
\tan 2x = \dfrac{{1 + \sqrt 5 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \arctan \dfrac{{1 – \sqrt 5 }}{2} + k\pi \\
2x = \arctan \dfrac{{1 + \sqrt 5 }}{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\arctan \dfrac{{1 – \sqrt 5 }}{2} + k\pi \\
x = \dfrac{1}{2}\arctan \dfrac{{1 + \sqrt 5 }}{2} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\sin 2x + 2\cos x = 0\\
\Leftrightarrow 2\sin x.\cos x + 2\cos x = 0\\
\Leftrightarrow 2\cos x.\left( {\sin x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)