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

Toán Lớp 11: Sin3x+sin2x=o Sinx=cos3x Sin4x+cosx=0

Tháng Chín 9, 2022 Toán học 1 Comment 0 views

Toán Lớp 11: Sin3x+sin2x=o
Sinx=cos3x
Sin4x+cosx=0

Comments ( 1 )

  1. danvanthu
    danvanthu
    9 Tháng Chín, 2022 at 11:33 sáng
    Reply
    sin3x+sin2x=0
    ⇔ sin3x=-sin2x
    ⇔ sin3x=sin-2x
    ⇔ $\left [\begin{array}{l} 3x=-2x+k2π \\ 3x=π+2x+k2π \end{array} \right.$
    ⇔ $\left [\begin{array}{l} 5x=k2π \\ x=π+k2π \end{array} \right.$
    ⇔ $\left [\begin{array}{l} x=\dfrac{k2π}{5} \\ x=π+k2π \end{array} \right. \ (k∈\mathbb{Z})$
    sinx=cos3x
    ⇔ sinx=sin(\frac{π}{2}-3x)
    ⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{2}-3x+k2π \\ x=π-\dfrac{π}{2}+3x+k2π \end{array} \right.$
    ⇔ $\left [\begin{array}{l} 4x=\dfrac{\pi}{2}+k2π \\ -2x=\dfrac{\pi}{2}+k2π \end{array} \right.$
    ⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{8}+\dfrac{kπ}{2} \\ x=-\dfrac{\pi}{4}-kπ \end{array} \right. \ (k∈\mathbb{Z})$
    sin4x+cosx=0
    ⇔ sin4x=-cosx
    ⇔ sin4x=cos(π-x)
    ⇔ sin4x=sin(x-\frac{π}{2})
    ⇔ $\left [\begin{array}{l} 4x=x-\dfrac{π}{2}+k2π \\ 4x=π-x+\dfrac{π}{2}+k2π \end{array} \right.$
    ⇔ $\left [\begin{array}{l} 3x=-\dfrac{π}{2}+k2π \\ 5x=\dfrac{3π}{2}+k2π \end{array} \right.$
    ⇔ $\left [\begin{array}{l} x=-\dfrac{π}{6}+\dfrac{k2π}{3} \\ x=\dfrac{3π}{10}+\dfrac{k2π}{5} \end{array} \right. \ (k∈\mathbb{Z})$

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

Thanh THương

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