## cos5xcos4x=cos3xcos2x

Question

cos5xcos4x=cos3xcos2x

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8 tháng 2020-10-30T04:30:28+00:00 2 Answers 63 views 0

1.  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.         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)