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

Toán Lớp 11: Giải phương trình sau 2sin^3 x+cos2x+cosx=0

Toán Lớp 11: Giải phương trình sau
2sin^3 x+cos2x+cosx=0

Comments ( 2 )

  1. Lời giải và giải thích chi tiết: \[\begin{array}{l} 2{\sin ^3}x – \cos 2x + cosx = 0\\ \Leftrightarrow 2sinx(1 – co{s^2}x) + (1 – 2{\cos ^2}x + \cos x) = 0\\ \Leftrightarrow 2\sin x(1 – \cos x)(1 + \cos x) – (2cosx + 1)(\cos x – 1) = 0\\ \Leftrightarrow (1 – \cos x)(2\sin x.\cos x + 2\sin x + 2\cos x + 1) = 0\\ \Leftrightarrow (1 – \cos x)({\sin ^2}x + {\cos ^2}x + 2\sin x\cos x + 2(\sin x + \cos x)) = 0\\ \Leftrightarrow (1 – \cos x)\left[ {{{(\sin x + \cos x)}^2} + 2(\sin x + \cos x)} \right] = 0\\ \Leftrightarrow (1 – \cos x)(\sin x + \cos x)(\sin x + \cos x + 2) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = 1\\ \sin x + \cos x = 0\\ \sin x + \cos x = – 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ \sin (x + \frac{\pi }{4}) = 0 \Leftrightarrow x = – \frac{\pi }{4} + k\pi \\ \sin (x + \frac{\pi }{4}) = – \sqrt 2 = > VN \end{array} \right. \end{array}\]

  2. 2sin^3x – cos2x + cosx = 0
    2sin^3x + (2sin^2x – 1) + cosx = 0
    (2sin^3x + 2sin^2x) – 1 + cosx = 0
    2sin^2x(sinx + 1) + cosx – 1 = 0
    2(1 – cos^2x)(sinx + 1) + (cosx – 1) = 0
    2(cos^2x – 1)(sinx + 1) – (cosx – 1) = 0
    2(cosx – 1)(cosx + 1)(sinx + 1) – (cosx – 1) = 0
    (cosx – 1)[2(cosx + 1)(sinx + 1) – 1] = 0
    (cosx – 1)(2sinxcosx + 2sinx + 2cosx + 1) = 0
    (cosx – 1)[(sinx + cosx)² + 2(sinx + cosx)] = 0
    (cosx – 1)(sinx + cosx)(sinx + cosx + 2) = 0
    Vì sinx + cosx + 2 ≠ 0 nên
    (cosx – 1)(sinx + cosx) = 0
    cosx – 1 = 0 hoặc sinx + cosx = 0
    (a) cosx – 1 = 0 ⇒ cosx = 1 ⇒ x = k 2π, (k ∈ Z)
    (b) sinx + cosx = 0 ⇒ tanx + 1 = 0 ⇒ tanx = -1 ⇒ x = 3π/4 + kπ, (k ∈ Z)
     

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