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

Toán Lớp 11: 2cosx+4sinx=3/cosx Giúp mik với mik đang cần gấp mik voice 5 sao

Toán Lớp 11: 2cosx+4sinx=3/cosx
Giúp mik với mik đang cần gấp mik voice 5 sao

Comments ( 1 )

  1. Giải đáp:
    \(\left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \\
    x = \arccos \dfrac{3}{{\sqrt {10} }} + k\pi 
    \end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
    Lời giải và giải thích chi tiết:
     ĐKXĐ:  \(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \)
    Ta có:
    \(\begin{array}{l}
    2\cos x + 4\sin x = \dfrac{3}{{\cos x}}\\
     \Leftrightarrow 2.{\cos ^2}x + 4\sin x.\cos x = 3\\
     \Leftrightarrow 2{\cos ^2}x + 4\sin x.\cos x = 3.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
     \Leftrightarrow 2{\cos ^2}x + 4\sin x.\cos x = 3\sin {x^2} + 3{\cos ^2}x\\
     \Leftrightarrow 3{\sin ^2}x – 4\sin x.\cos x + {\cos ^2}x = 0\\
     \Leftrightarrow \left( {3{{\sin }^2}x – 3\sin x.\cos x} \right) + \left( { – \sin x.\cos x + {{\cos }^2}x} \right) = 0\\
     \Leftrightarrow 3\sin x.\left( {\sin x – \cos x} \right) – \cos x.\left( {\sin x – \cos x} \right) = 0\\
     \Leftrightarrow \left( {\sin x – \cos x} \right).\left( {3\sin x – \cos x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x – \cos x = 0\\
    3\sin x – \cos x = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \dfrac{1}{{\sqrt 2 }}.\sin x – \dfrac{1}{{\sqrt 2 }}.\cos x = 0\\
    \dfrac{3}{{\sqrt {10} }}\sin x – \dfrac{1}{{\sqrt {10} }}.\cos x = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x.\cos \dfrac{\pi }{4} – \cos x.\sin \dfrac{\pi }{4} = 0\\
    \sin x.\cos \left( {\arccos \dfrac{3}{{\sqrt {10} }}} \right) – \cos x.\sin \left( {\arccos \dfrac{3}{{\sqrt {10} }}} \right) = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin \left( {x – \dfrac{\pi }{4}} \right) = 0\\
    \sin \left( {x – \arccos \dfrac{3}{{\sqrt {10} }}} \right) = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – \dfrac{\pi }{4} = k\pi \\
    x – \arccos \dfrac{3}{{\sqrt {10} }} = k\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \\
    x = \arccos \dfrac{3}{{\sqrt {10} }} + k\pi 
    \end{array} \right.\,\,\,\,\left( {k \in Z} \right)
    \end{array}\)

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