Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

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

Toán Lớp 11: mn cho em hỏi vài vâu ạ 1) cosx = cos2x 2) cos ( x – 2π/5 ) = cos2x 3) tan ( x – π/4 ) = tan2x cảm ơn mọi người ????

Toán Lớp 11: mn cho em hỏi vài vâu ạ
1) cosx = cos2x
2) cos ( x – 2π/5 ) = cos2x
3) tan ( x – π/4 ) = tan2x
cảm ơn mọi người ????

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
     1) cos\ x=cos\ 2x
    ⇔ \(\left[ \begin{array}{l}x=2x+k2\pi\ (k \in \mathbb{Z})\\x=-2x+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}-x=k2\pi\ (k \in \mathbb{Z})\\3x=k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-k2\pi\ (k \in \mathbb{Z})\\x=k\dfrac{2\pi}{3}\ (k \in \mathbb{Z})\end{array} \right.\) 
    Vậy S={-k2\pi\ (k \in \mathbb{Z});k\frac{2\pi}{3}\ (k \in \mathbb{Z})}
    2) cos\ (x-\frac{2\pi}{5})=cos\ 2x
    ⇔ \(\left[ \begin{array}{l}x-\dfrac{2\pi}{5}=2x+k2\pi\ (k \in \mathbb{Z})\\x-\dfrac{2\pi}{5}=-2x+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}-x=\dfrac{2\pi}{5}+k2\pi\ (k \in \mathbb{Z})\\3x=\dfrac{2\pi}{5}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-\dfrac{2\pi}{5}-k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{2\pi}{15}+k\dfrac{2\pi}{3}\ (k \in \mathbb{Z})\end{array} \right.\) 
    Vậy S={-\frac{2\pi}{5}-k2\pi\ (k \in \mathbb{Z});\frac{2\pi}{15}+k\frac{2\pi}{3}\ (k \in \mathbb{Z})}
    3) tan\ (x-\frac{\pi}{4})=tan\ 2x
    ⇔ x-\frac{\pi}{4}=2x+k\pi\ (k \in \mathbb{Z})
    ⇔ -x=\frac{\pi}{4}+k\pi\ (k \in \mathbb{Z})
    ⇔ x=-\frac{\pi}{4}-k\pi\ (k \in \mathbb{Z})
    Vậy S={-\frac{\pi}{4}-k\pi\ (k \in \mathbb{Z})}

Leave a reply

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

About Thu Ánh