Toán Lớp 11: Giúp mình với ạ, gấp ạ
Tìm nghiệm của phương trình:
cos(π/2 + 2x) – √3 cos(π – 2x) = 1
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Toán Lớp 11: Giúp mình với ạ, gấp ạ
Tìm nghiệm của phương trình:
cos(π/2 + 2x) – √3 cos(π – 2x) = 1
Comments ( 2 )
\quad \cos\left(\dfrac{\pi}{2} + 2x\right) – \sqrt3\cos\left(\pi – 2x\right) = 1\\
\Leftrightarrow -\sin2x + \sqrt3\cos2x = 1\\
\Leftrightarrow \dfrac{\sqrt3}{2}\cos2x – \dfrac12\sin2x = \dfrac12\\
\Leftrightarrow \cos2x.\cos\dfrac{\pi}{6} – \sin2x.\sin\dfrac{\pi}{6} = \cos\dfrac{\pi}{3}\\
\Leftrightarrow \cos\left(2x + \dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{3}\\
\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{6} = \dfrac{\pi}{3} +k2\pi\\2x + \dfrac{\pi}{6} = – \dfrac{\pi}{3} +k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\pi\\x = – \dfrac{\pi}{4} +k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{- \dfrac{\pi}{4} +k\pi;\ \dfrac{\pi}{12} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
x = \dfrac{\pi }{{12}} + k\pi \\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
\cos \left( {\dfrac{\pi }{2} + 2x} \right) – \sqrt 3 \cos \left( {\pi – 2x} \right) = 1\\
\Leftrightarrow \sin \left[ {\dfrac{\pi }{2} – \left( {\dfrac{\pi }{2} + 2x} \right)} \right] – \sqrt 3 .\left[ { – \cos \left( {\pi – \left( {\pi – 2x} \right)} \right)} \right] = 1\\
\Leftrightarrow \sin \left( { – 2x} \right) + \sqrt 3 .\cos 2x = 1\\
\Leftrightarrow – \sin 2x + \sqrt 3 \cos 2x = 1\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos 2x – \dfrac{1}{2}\sin 2x = \dfrac{1}{2}\\
\Leftrightarrow \cos 2x.\cos \dfrac{\pi }{6} – \sin 2x.\sin \dfrac{\pi }{6} = \dfrac{1}{2}\\
\Leftrightarrow \cos \left( {2x + \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{\pi }{6} = – \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
2x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)