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

Toán Lớp 11: Giải phương trình cos(x-2) =sin 3(x-π/2)

Toán Lớp 11: Giải phương trình
cos(x-2) =sin 3(x-π/2)

Comments ( 2 )

  1. $cos(x-2) =sin 3\bigg(x-\dfrac{\pi}{2}\bigg)$
    $⇔cos(x-2) =sin \bigg(3x-\dfrac{3\pi}{2}\bigg)$
    $⇔cos(x-2) =cos \bigg[\dfrac{\pi}{2}-(3x-\dfrac{3\pi}{2})\bigg]$
    $⇔cos(x-2) =cos \bigg[\dfrac{\pi}{2}-(3x-\dfrac{3\pi}{2})\bigg]$
    $⇔cos(x-2)=cos(2\pi-3x)$
    $⇔$\(\left[ \begin{array}{l}x-2=2\pi-3x+k2\pi\\x-2=3x-2\pi+k2\pi\end{array} \right.\) 
    $⇔$\(\left[ \begin{array}{l}4x=2+2\pi+k2\pi\\-2x=2-2\pi+k2\pi\end{array} \right.\) 
    $⇔$\(\left[ \begin{array}{l}x=\dfrac{1}{2}+\dfrac{\pi}{2}+\dfrac{k\pi}{2}\\x=-1+\pi-k\pi\end{array} \right.\) $(k∈Z)$

  2. $\begin{array}{l}
    \cos \left( {x – 2} \right) = \sin 3\left( {x – \dfrac{\pi }{2}} \right)\\
     \Leftrightarrow \cos \left( {x – 2} \right) = \sin \left( {3x – \dfrac{{3\pi }}{2}} \right)\\
     \Leftrightarrow \cos \left( {x – 2} \right) = \sin \left( {\dfrac{\pi }{2} +\left( { – 2\pi  + 3x} \right)} \right)\\
     \Leftrightarrow \cos \left( {x – 2} \right) = \cos \left( {3x – 2\pi } \right)\\
     \Leftrightarrow \cos \left( {x – 2} \right) = \cos 3x\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 2 = 3x + k2\pi \\
    x – 2 =  – 3x + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x =  – 1 – k\pi \\
    x = \dfrac{1}{2} + \dfrac{{k\pi }}{2}
    \end{array} \right.\left( {k \in \mathbb{Z}} \right)
    \end{array}$
     

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