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

Toán Lớp 11: cos^2x+1/cos^2x+cosx-1/cosx-7/4=0

Toán Lớp 11: cos^2x+1/cos^2x+cosx-1/cosx-7/4=0

Comments ( 1 )

  1. $\begin{array}{l}
    ĐK:\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
    {\cos ^2}x + \dfrac{1}{{{{\cos }^2}x}} + \cos x – \dfrac{1}{{\cos x}} – \dfrac{7}{4} = 0\\
     \Leftrightarrow {\left( {\cos x – \dfrac{1}{{\cos x}}} \right)^2} + \left( {\cos x – \dfrac{1}{{\cos x}}} \right) + 2 – \dfrac{7}{4} = 0\\
     \Leftrightarrow {\left( {\cos x – \dfrac{1}{{\cos x}}} \right)^2} + \left( {\cos x – \dfrac{1}{{\cos x}}} \right) + \dfrac{1}{4} = 0\\
     \Leftrightarrow {\left( {\cos x – \dfrac{1}{{\cos x}} + \dfrac{1}{2}} \right)^2} = 0\\
     \Leftrightarrow \cos x – \dfrac{1}{{\cos x}} + \dfrac{1}{2} = 0\\
     \Leftrightarrow 2{\cos ^2}x – 2 + \cos x = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos x = \dfrac{{ – 1 + \sqrt {17} }}{4}\\
    \cos x = \dfrac{{ – 1 – \sqrt {17} }}{4}(L)
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = \arccos \dfrac{{ – 1 + \sqrt {17} }}{4} + k2\pi \\
    x =  – \arccos \dfrac{{ – 1 + \sqrt {17} }}{4} + k2\pi 
    \end{array} \right.\left( {k \in \mathbb{Z}} \right)
    \end{array}$

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