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

Toán Lớp 11: Giúp em với ạ tan (x + π/3 ) – cot3x = 0

Toán Lớp 11: Giúp em với ạ
tan (x + π/3 ) – cot3x = 0

Comments ( 2 )

  1. tan(x+$\pi$/3) – cot3x = 0
    ⇔tan(x+$\pi$/3) = cot3x
    ⇔tan(x+$\pi$/3) = tan($\pi$/2-3x)
    ⇔x+$\pi$/3 = $\pi$/2 – 3x + k$\pi$ (k∈Z)
    ⇔4x = $\pi$/6 + k$\pi$ (k∈Z)
    ⇔x=$\pi$/24 +k$\pi$/4 (k∈Z)
     

  2. Giải đáp: x=\frac{π}{24}+\frac{kπ}{4} \ (k∈\mathbb{Z})
    Giải:
    tan(x+\frac{π}{3})-cot3x=0
    Đkxđ:
    *) cos(x+\frac{π}{3}) \ne 0
    ⇔ x+\frac{π}{3} \ne \frac{π}{2}+kπ
    ⇔ x \ne \frac{π}{6}+kπ \ (k∈\mathbb{Z})
    *) sin3x \ne 0
    ⇔ 3x \ne kπ
    ⇔ x \ne \frac{kπ}{3} \ (k∈\mathbb{Z})
    Ta có:
    tan(x+\frac{π}{3})-cot3x=0
    ⇔ tan(x+\frac{π}{3})=cot3x
    ⇔ tan(x+\frac{π}{3})=tan(\frac{π}{2}-3x)
    ⇔ x+\frac{π}{3}=\frac{π}{2}-3x+kπ
    ⇔ 4x=\frac{π}{6}+kπ
    ⇔ x=\frac{π}{24}+\frac{kπ}{4} \ (k∈\mathbb{Z}) (thỏa)
    Vậy x=\frac{π}{24}+\frac{kπ}{4} \ (k∈\mathbb{Z})

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