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

Toán Lớp 11: 1) 2Sin ²x + Sinx – 1 = 0 2) 3Cos2x – 2Cosx – 8 = 0 3) √3Sinx + Cosx = 1 4)Sin4x – Sin6x = √3(Cos6x + Cos4x) giúp mình với, mình cần

Home/ Toán Lớp 11: 1) 2Sin ²x + Sinx – 1 = 0 2) 3Cos2x – 2Cosx – 8 = 0 3) √3Sinx + Cosx = 1 4)Sin4x – Sin6x = √3(Cos6x + Cos4x) giúp mình với, mình cần

Toán Lớp 11: 1) 2Sin ²x + Sinx – 1 = 0 2) 3Cos2x – 2Cosx – 8 = 0 3) √3Sinx + Cosx = 1 4)Sin4x – Sin6x = √3(Cos6x + Cos4x) giúp mình với, mình cần

Tháng Một 26, 2023 Toán học 2 Comments 0 views

Toán Lớp 11: 1) 2Sin ²x + Sinx – 1 = 0
2) 3Cos2x – 2Cosx – 8 = 0
3) √3Sinx + Cosx = 1
4)Sin4x – Sin6x = √3(Cos6x + Cos4x)
giúp mình với, mình cần gấp ạ

Comments ( 2 )

  1. dantam45
    dantam45
    26 Tháng Một, 2023 at 3:33 sáng
    Reply
    1)
    $2\sin^2x+\sin x-1=0$
    $\to \left[ \begin{array}{l}\sin x=-1 \\ \sin x=\dfrac{1}{2}\end{array} \right.$
    $\to \left[ \begin{array}{l}x=\dfrac{-\pi}{2}+k2\pi \\x=\dfrac{\pi}{6}+k2\pi\\ x=\dfrac{5\pi}{6}+k2\pi \end{array} \right.$
    2)
    $3\cos2x-2\cos x-8=0$
    $\to 3(2\cos^2x-1)-2\cos x-8=0$
    $\to 6\cos^2x-2\cos x-11=0$
    $\to \cos x=\dfrac{1\pm\sqrt{67}}{6}\notin[-1;1]$ (loại)
    Vậy PT vô nghiệm
    3)
    $\sqrt3\sin x+\cos x=1$
    $\to 2\sin\left(x+\dfrac{\pi}{6}\right)=1$ 
    $\to \sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}$
    $\to \left[ \begin{array}{l}x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi \\x+\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi \end{array} \right.$
    $\to \left[ \begin{array}{l}x=k2\pi \\x=\dfrac{2\pi}{3}+k2\pi \end{array} \right.$
    4)
    $\sin4x-\sin6x=\sqrt3(\cos6x+\cos4x)$
    $\to \sin4x-\sqrt3\cos4x=\sin6x+\sqrt3\cos6x$
    $\to \dfrac{1}{2}\sin4x-\dfrac{\sqrt3}{2}\cos4x=\dfrac{1}{2}\sin6x+\dfrac{\sqrt3}{2}\cos6x$
    $\to \sin\left(4x-\dfrac{\pi}{3}\right)=\sin\left( 6x+\dfrac{\pi}{3}\right)$
    $\to \left[ \begin{array}{l}6x+\dfrac{\pi}{3}=4x-\dfrac{\pi}{3}+k2\pi \\6x+\dfrac{\pi}{3}=\pi-4x+\dfrac{\pi}{3}+k2\pi \end{array} \right.$
    $\to \left[ \begin{array}{l}x=\dfrac{-\pi}{3}+k\pi \\x=\dfrac{\pi}{10}+\dfrac{k\pi}{5} \end{array} \right.$

  2. maianhnhiquynh
    maianhnhiquynh
    26 Tháng Một, 2023 at 3:33 sáng
    Reply
    CHÚC BẠN HỌC TỐT!!! 
    Lời giải và giải thích chi tiết:
     

    toan-lop-11-1-2sin-sin-1-0-2-3cos2-2cos-8-0-3-3sin-cos-1-4-sin4-sin6-3-cos6-cos4-giup-minh-voi-m

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

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