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

Toán Lớp 11: Giúp mình với, mình cần gấp:(( cot (2x-π/4)=cot (x+π/3)

Toán Lớp 11: Giúp mình với, mình cần gấp:((
cot (2x-π/4)=cot (x+π/3)

Comments ( 1 )

  1. Giải đáp:
    \(x = \dfrac{{7\pi }}{{12}} + k\pi \) 
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    DK:\left\{ \begin{array}{l}
    \sin \left( {2x – \dfrac{\pi }{4}} \right) \ne 0\\
    \sin \left( {x + \dfrac{\pi }{3}} \right) \ne 0
    \end{array} \right. \to \left\{ \begin{array}{l}
    2x – \dfrac{\pi }{4} \ne k\pi \\
    x + \dfrac{\pi }{3} \ne k\pi 
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    x \ne \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
    x \ne  – \dfrac{\pi }{3} + k\pi 
    \end{array} \right.\left( {k \in Z} \right)\\
    \cot \left( {2x – \dfrac{\pi }{4}} \right) = \cot \left( {x + \dfrac{\pi }{3}} \right)\\
     \to \tan \left( {2x – \dfrac{\pi }{4}} \right) = \tan \left( {x + \dfrac{\pi }{3}} \right)\\
     \to 2x – \dfrac{\pi }{4} = x + \dfrac{\pi }{3} + k\pi \\
     \to x = \dfrac{{7\pi }}{{12}} + k\pi \left( {k \in Z} \right)
    \end{array}\)

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