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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: ( √3 sin X – cos X ) ( 2 – √3sin X + cos X ) = 0

Toán Lớp 11: ( √3 sin X – cos X ) ( 2 – √3sin X + cos X ) = 0

Comments ( 1 )

  1. Giải đáp:
    \[\left[ \begin{array}{l}
    x = \dfrac{\pi }{6} + k\pi \\
    x = \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
    Lời giải và giải thích chi tiết:
     Ta có:
    \(\begin{array}{l}
    \left( {\sqrt 3 \sin x – \cos x} \right)\left( {2 – \sqrt 3 \sin x + \cos x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sqrt 3 \sin x – \cos x = 0\\
    2 – \sqrt 3 \sin x + \cos x = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sqrt 3 \sin x – \cos x = 0\\
    \sqrt 3 \sin x – \cos x = 2
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \dfrac{{\sqrt 3 }}{2}\sin x – \dfrac{1}{2}\cos x = 0\\
    \dfrac{{\sqrt 3 }}{2}\sin x – \dfrac{1}{2}\cos x = 1
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x.\cos \dfrac{\pi }{6} – \cos x.\sin \dfrac{\pi }{6} = 0\\
    \sin x.\cos \dfrac{\pi }{6} – \cos x.\sin \dfrac{\pi }{6} = 1
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin \left( {x – \dfrac{\pi }{6}} \right) = 0\\
    \sin \left( {x – \dfrac{\pi }{6}} \right) = 1
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – \dfrac{\pi }{6} = k\pi \\
    x – \dfrac{\pi }{6} = \dfrac{\pi }{2} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{6} + k\pi \\
    x = \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\,\,\,\,\left( {k \in Z} \right)
    \end{array}\)

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222-9+11+12:2*14+14 = ? ( )