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: 1. sin3x = sinx 2. sinx – cos2x =0 3. sin2x = cos3x 4. sin4x + cos5x=0

Home/ Toán Lớp 11: 1. sin3x = sinx 2. sinx – cos2x =0 3. sin2x = cos3x 4. sin4x + cos5x=0

Toán Lớp 11: 1. sin3x = sinx 2. sinx – cos2x =0 3. sin2x = cos3x 4. sin4x + cos5x=0

Tháng Hai 15, 2023 Toán học 2 Comments 0 views

Toán Lớp 11: 1. sin3x = sinx
2. sinx – cos2x =0
3. sin2x = cos3x
4. sin4x + cos5x=0

Comments ( 2 )

  1. tongtung
    tongtung
    15 Tháng Hai, 2023 at 1:06 sáng
    Reply
    1. sin3x = sinx
    ⇔ \(\left[ \begin{array}{l}3x=x+k2π\\3x=π-x+k2π\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}2x=k2π\\4x=π+k2π\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=kπ\\x=π/4+kπ/2\end{array} \right.\) 
    ⇔ x=π/4 +kπ/2 (k∈Z)
    ____________________________________________________
    2. sinx – cos2x =0
    ⇔ sinx = cos2x
    ⇔ sinx= sin(π/2  – x)
    ⇔ \(\left[ \begin{array}{l}x=π/2-x+k2π\\x=π-π/2+x+k2π\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=π/4+kπ(k∈Z)\\x=π-π/2+x+k2π(loại)\end{array} \right.\) 
    ________________________________
    3. sin2x = cos3x
    ⇔ sin2x = sin(π/2 – 3x)
    ⇔ \(\left[ \begin{array}{l}2x=π/3-3x+k2π\\2x=π-π/2+3x+k2π\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=π/10+k2π/5\\x=-π/2-k2π\end{array} \right.\) (k∈Z)
    ______________________________________________
    4. sin4x + cos5x=0
    ⇔ cos5x = sin(-4x)
    ⇔ cos5x = sin(π/2 +4x)
    ⇔ \(\left[ \begin{array}{l}5x=π/2+4x+k2π\\5x=-π/2-4x+k2π\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=π/2+k2π\\x=-π/18+k2π/9\end{array} \right.\) (k∈Z)

  2. hatuanlan
    hatuanlan
    15 Tháng Hai, 2023 at 1:06 sáng
    Reply
    ~rai~
    \(1.\sin3x=\sin x\\\Leftrightarrow \left[\begin{array}{I}3x=x+k2\pi\\3x=\pi-x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=k2\pi\\4x=\pi+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=k\pi\\x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\end{array}\right.\\\Leftrightarrow x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\2)\sin x-\cos2x=0\\\Leftrightarrow \sin x=\cos2x\\\Leftrightarrow \sin x=\sin\left(\dfrac{\pi}{2}-x\right)\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}-x+k2\pi\\x=\pi-\dfrac{\pi}{2}+x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=\dfrac{\pi}{2}+k2\pi\\0=\dfrac{\pi}{2}+k2\pi.\text{(vô lí)}\end{array}\right.\\\Leftrightarrow x=\dfrac{\pi}{4}+k\pi.(k\in\mathbb{Z})\\3.\sin2x=\cos3x\\\Leftrightarrow \sin2x=\sin\left(\dfrac{\pi}{2}-3x\right)\\\Leftrightarrow \left[\begin{array}{I}2x=\dfrac{\pi}{3}-3x+k2\pi\\2x=\pi-\dfrac{\pi}{2}+3x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}5x=\dfrac{\pi}{2}+k2\pi\\-x=\dfrac{\pi}{2}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{10}+k\dfrac{2\pi}{5}\\x=-\dfrac{\pi}{2}-k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\4.\sin4x+\cos5x=0\\\Leftrightarrow \cos5x=-sin4x\\\Leftrightarrow \cos5x=\sin(-4x)\\\Leftrightarrow \cos5x=\sin\left(\dfrac{\pi}{2}+4x\right)\\\Leftrightarrow \left[\begin{array}{I}5x=\dfrac{\pi}{2}+4x+k2\pi\\5x=-\dfrac{\pi}{2}-4x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}+k2\pi\\9x=-\dfrac{\pi}{2}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{18}+k\dfrac{2\pi}{9}.\end{array}\right.\quad(k\in\mathbb{Z})\)

Leave a reply

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

Kiều Nguyệt

About Kiều Nguyệt