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

Toán Lớp 11: a) sin^2x-3sinx+2=0 b) 5cosx-2sin2x=0

Toán Lớp 11: a) sin^2x-3sinx+2=0
b) 5cosx-2sin2x=0

Comments ( 1 )

  1. Giải đáp:
    \(\begin{array}{l}
    a,\\
    x = \dfrac{\pi }{2} + k2\pi \,\,\,\left( {k \in Z} \right)\\
    b,\\
    x = \dfrac{\pi }{2} + k\pi \,\,\,\,\left( {k \in Z} \right)
    \end{array}\)
    Lời giải và giải thích chi tiết:
     Ta có:
    \(\begin{array}{l}
    a,\\
    {\sin ^2}x – 3\sin x + 2 = 0\\
     \Leftrightarrow \left( {{{\sin }^2}x – \sin x} \right) + \left( { – 2\sin x + 2} \right) = 0\\
     \Leftrightarrow \sin x.\left( {\sin x – 1} \right) – 2.\left( {\sin x – 1} \right) = 0\\
     \Leftrightarrow \left( {\sin x – 1} \right)\left( {\sin x – 2} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x – 1 = 0\\
    \sin x – 2 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 1\\
    \sin x = 2
    \end{array} \right.\\
     – 1 \le \sin x \le 1 \Rightarrow \sin x = 1 \Rightarrow x = \dfrac{\pi }{2} + k2\pi \,\,\,\left( {k \in Z} \right)\\
    b,\\
    5\cos x – 2\sin 2x = 0\\
     \Leftrightarrow 5\cos x – 2.2\sin x.\cos x = 0\\
     \Leftrightarrow 5\cos x – 4\sin x.\cos x = 0\\
     \Leftrightarrow \cos x.\left( {5 – 4\sin x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos x = 0\\
    5 – 4\sin x = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    \cos x = 0\\
    \sin x = \dfrac{5}{4}
    \end{array} \right.\\
     – 1 \le \sin x \le 1 \Rightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\,\,\left( {k \in Z} \right)
    \end{array}\)

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

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