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

Toán Lớp 11: giải phương trình: sin (3x+1) =1/2

Toán Lớp 11: giải phương trình: sin (3x+1) =1/2

Comments ( 2 )

  1. $sin(3x+1)=\dfrac{1}{2}$
    $⇔sin(3x+1)=sin(\dfrac{\pi}{6})$
    $⇔$\(\left[ \begin{array}{l}3x+1=\dfrac{\pi}{6}+k2\pi\\3x+1=\pi-\dfrac{\pi}{6}+k2\pi\end{array} \right.\) 
    $⇔$\(\left[ \begin{array}{l}3x=-1+\dfrac{\pi}{6}+k2\pi\\3x=-1+\dfrac{5\pi}{6}+k2\pi\end{array} \right.\) 
    $⇔$\(\left[ \begin{array}{l}x=-\dfrac{1}{3}+\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=-\dfrac{1}{3}+\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{array} \right.\) 

  2. Giải đáp + Lời giải và giải thích chi tiết:
    sin (3x+1) =1/2
    <=>sin(3x+1)=sin(pi/6)
    <=>\(\left[ \begin{array}{l}3x+1=\dfrac{\pi}{6}+k2\pi\\3x+1=\pi-\dfrac{\pi}{6}+k2\pi\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}3x=\dfrac{\pi}{6}-1+k2\pi\\3x=\dfrac{5\pi}{6}-1+k2\pi\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=\dfrac{\pi}{18}-\dfrac{1}{3}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}-\dfrac{1}{3}+\dfrac{k2\pi}{3}\end{array} \right.\)(kinZZ)

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