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

Toán Lớp 11: 1, sin2x+sinx=0 2, sinx+cos2x-1=0 3, cosx + cos2x+1=0 4, sin^2 3x+sin^2 4x=1 5, sin(pi/6+x)+cos(pi/3+2x)=1

Toán Lớp 11: 1, sin2x+sinx=0
2, sinx+cos2x-1=0
3, cosx + cos2x+1=0
4, sin^2 3x+sin^2 4x=1
5, sin(pi/6+x)+cos(pi/3+2x)=1

Comments ( 2 )

  1. CHÚC BẠN HỌC TỐT!!! 
    Lời giải và giải thích chi tiết:
     

    toan-lop-11-1-sin2-sin-0-2-sin-cos2-1-0-3-cos-cos2-1-0-4-sin-2-3-sin-2-4-1-5-sin-pi-6-cos-pi-3-2

  2. Giải đáp:
    \(\begin{array}{l}
    1,\\
    \left[ \begin{array}{l}
    x = k\pi \\
    x = \dfrac{{2\pi }}{3} + k2\pi \\
    x =  – \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
    2,\\
    \left[ \begin{array}{l}
    x = k\pi \\
    x = \dfrac{\pi }{6} + k2\pi \\
    x = \dfrac{{5\pi }}{6} + k2\pi 
    \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
    3,\\
    \left[ \begin{array}{l}
    x = \dfrac{\pi }{2} + k\pi \\
    x = \dfrac{{2\pi }}{3} + k2\pi \\
    x =  – \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
    4,\\
    \left[ \begin{array}{l}
    x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
    x = \dfrac{\pi }{2} + k\pi \\
    x = \dfrac{{3\pi }}{{14}} + \dfrac{{k2\pi }}{7}
    \end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
    5,\\
    \left[ \begin{array}{l}
    x =  – \dfrac{\pi }{6} + k\pi \\
    x = k2\pi \\
    x = \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\,\,\,\,\left( {k \in Z} \right)
    \end{array}\) 
    Lời giải và giải thích chi tiết:
     Ta có:
    \(\begin{array}{l}
    1,\\
    \sin 2x + \sin x = 0\\
     \Leftrightarrow 2\sin x.\cos x + \sin x = 0\\
     \Leftrightarrow \sin x.\left( {2\cos x + 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 0\\
    2\cos x + 1 = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 0\\
    \cos x =  – \dfrac{1}{2}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 0\\
    \cos x = \cos \dfrac{{2\pi }}{3}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = k\pi \\
    x = \dfrac{{2\pi }}{3} + k2\pi \\
    x =  – \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
    2,\\
    \sin x + \cos 2x – 1 = 0\\
     \Leftrightarrow \sin x + \left( {1 – 2{{\sin }^2}x} \right) – 1 = 0\\
     \Leftrightarrow \sin x + 1 – 2{\sin ^2}x – 1 = 0\\
     \Leftrightarrow \sin x – 2{\sin ^2}x = 0\\
     \Leftrightarrow \sin x.\left( {1 – 2\sin x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 0\\
    1 – 2\sin x = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 0\\
    \sin x = \dfrac{1}{2}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 0\\
    \sin x = \sin \dfrac{\pi }{6}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = k\pi \\
    x = \dfrac{\pi }{6} + k2\pi \\
    x = \left( {\pi  – \dfrac{\pi }{6}} \right) + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = k\pi \\
    x = \dfrac{\pi }{6} + k2\pi \\
    x = \dfrac{{5\pi }}{6} + k2\pi 
    \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
    3,\\
    \cos x + \cos 2x + 1 = 0\\
     \Leftrightarrow \cos x + \left( {2{{\cos }^2}x – 1} \right) + 1 = 0\\
     \Leftrightarrow \cos x + 2{\cos ^2}x – 1 + 1 = 0\\
     \Leftrightarrow \cos x + 2{\cos ^2}x = 0\\
     \Leftrightarrow \cos x.\left( {1 + 2\cos x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos x = 0\\
    1 + 2\cos x = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos x = 0\\
    \cos x =  – \dfrac{1}{2}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos x = 0\\
    \cos x = \cos \dfrac{{2\pi }}{3}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{2} + k\pi \\
    x = \dfrac{{2\pi }}{3} + k2\pi \\
    x =  – \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
    4,\\
    {\sin ^2}3x + {\sin ^2}4x = 1\\
     \Leftrightarrow {\sin ^2}3x + \left( {1 – {{\cos }^2}4x} \right) = 1\\
     \Leftrightarrow {\sin ^2}3x + 1 – {\cos ^2}4x – 1 = 0\\
     \Leftrightarrow {\sin ^2}3x – {\cos ^2}4x = 0\\
     \Leftrightarrow \left( {\sin 3x – \cos x4x} \right).\left( {\sin 3x + \cos 4x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin 3x – \cos 4x = 0\\
    \sin 3x + \cos 4x = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin 3x = \cos 4x\\
    \sin 3x =  – \cos 4x
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin 3x = \sin \left( {\dfrac{\pi }{2} – 4x} \right)\\
    \sin 3x =  – \sin \left( {\dfrac{\pi }{2} – 4x} \right)
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin 3x = \sin \left( {\dfrac{\pi }{2} – 4x} \right)\\
    \sin 3x = \sin \left( {4x – \dfrac{\pi }{2}} \right)
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    3x = \dfrac{\pi }{2} – 4x + k2\pi \\
    3x = \pi  – \left( {\dfrac{\pi }{2} – 4x} \right) + k2\pi \\
    3x = 4x – \dfrac{\pi }{2} + k2\pi \\
    3x = \pi  – \left( {4x – \dfrac{\pi }{2}} \right) + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    3x = \dfrac{\pi }{2} – 4x + k2\pi \\
    3x = \dfrac{\pi }{2} + 4x + k2\pi \\
    3x = 4x – \dfrac{\pi }{2} + k2\pi \\
    3x = \dfrac{{3\pi }}{2} – 4x + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    7x = \dfrac{\pi }{2} + k2\pi \\
     – x = \dfrac{\pi }{2} + k2\pi \\
     – x =  – \dfrac{\pi }{2} + k2\pi \\
    7x = \dfrac{{3\pi }}{2} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
    x =  – \dfrac{\pi }{2} + k2\pi \\
    x = \dfrac{\pi }{2} + k2\pi \\
    x = \dfrac{{3\pi }}{{14}} + \dfrac{{k2\pi }}{7}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}\\
    x = \dfrac{\pi }{2} + k\pi \\
    x = \dfrac{{3\pi }}{{14}} + \dfrac{{k2\pi }}{7}
    \end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
    5,\\
    \sin \left( {\dfrac{\pi }{6} + x} \right) + \cos \left( {\dfrac{\pi }{3} + 2x} \right) = 1\\
     \Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) + \cos \left[ {2.\left( {\dfrac{\pi }{6} + x} \right)} \right] = 1\\
     \Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) + \left[ {1 – 2{{\sin }^2}\left( {\dfrac{\pi }{6} + x} \right)} \right] = 1\\
     \Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) + 1 – 2{\sin ^2}\left( {\dfrac{\pi }{6} + x} \right) = 1\\
     \Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right) – 2{\sin ^2}\left( {\dfrac{\pi }{6} + x} \right) = 0\\
     \Leftrightarrow \sin \left( {\dfrac{\pi }{6} + x} \right)\left( {1 – 2\sin \left( {\dfrac{\pi }{6} + x} \right)} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin \left( {\dfrac{\pi }{6} + x} \right) = 0\\
    1 – 2\sin \left( {\dfrac{\pi }{6} + x} \right) = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin \left( {\dfrac{\pi }{6} + x} \right) = 0\\
    \sin \left( {\dfrac{\pi }{6} + x} \right) = \dfrac{1}{2}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin \left( {\dfrac{\pi }{6} + x} \right) = 0\\
    \sin \left( {\dfrac{\pi }{6} + x} \right) = \sin \dfrac{\pi }{6}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \dfrac{\pi }{6} + x = k\pi \\
    \dfrac{\pi }{6} + x = \dfrac{\pi }{6} + k2\pi \\
    \dfrac{\pi }{6} + x = \left( {\pi  – \dfrac{\pi }{6}} \right) + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x =  – \dfrac{\pi }{6} + k\pi \\
    x = k2\pi \\
    x = \pi  – \dfrac{\pi }{6} – \dfrac{\pi }{6} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x =  – \dfrac{\pi }{6} + k\pi \\
    x = k2\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 = ? ( )