cos5xcos4x=cos3xcos2x

Question

cos5xcos4x=cos3xcos2x

in progress 0
Eirian 4 years 2020-10-30T04:30:28+00:00 2 Answers 98 views 0

Answers ( )

  1. Bo
    0
    2020-10-30T04:31:37+00:00
    Reply

    ` cos5xcos4x = cos3xcos2x `

    ` <=> 1/2 . (cosx + cos9x) = 1/2 . (cosx + cos5x) `

    ` <=> \frac{cosx}{2} + \frac{cos9x}{2} = \frac{cosx}{2} + \frac{cos5x}{2} `

    ` <=> \frac{cos9x}{2} = \frac{cos5x}{2} `

    ` <=> cos9x – cos5x = 0 `

    ` <=> -2sin7xsin2x = 0 `

    ` <=> sin7xsin2x = 0 `

    ` <=> ` \(\left[ \begin{array}{l}sin7x=0\\sin5x=0\end{array} \right.\) 

    ` <=> ` \(\left[ \begin{array}{l}x=\frac{kπ}{7}\\x=\frac{kπ}{2}\end{array} \right.\) `(k ∈ Z)`

  2. Jezebel
    0
    2020-10-30T04:31:55+00:00
    Reply

            cos5x . cos4x = cos3x . cos2x
    ⇔ 1/2.[cos(5x-4x)+cos(5x+4x)] = 1/2.[cos(3x-2x)+cos(3x+2x)]

    ⇔ 1/2.[cosx + cos9x]                 = 1/2.[cosx + cos5x]

    ⇔ 1/2.cosx + 1/2.cos9x             = 1/2.cosx + 1/2.cos5x

     đơn giản 1/2

    ⇔ cosx + cos9x = cosx + cos5x

    ⇔ cosx + cos9x – cosx – cos5x=0

    ⇔ cos9x – cos5x = 0

    ⇔ -2.sin7x.sin2x = 0

    ⇔ sin7x.sin2x = 0

    ⇔ \(\left[ \begin{array}{l}sin7x=0\\sin2x=0\end{array} \right.\) 

    ⇔ \(\left[ \begin{array}{l}x=k.π/7\\x=k.π/2\end{array} \right.\)  (k∈Z)

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )