Giải pt: cosx + cos2x + cos3x + cos4x = 0

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1. Ta có

   $\cos x + \cos(2x) + \cos(3x) + \cos(4x) = 0$

   $<-> \cos x + \cos(3x) + \cos(2x) + \cos(4x) = 0$

   $<-> 2\cos(2x) \cos x + 2 \cos(3x) \cos x = 0$

   $<-> \cos x[\cos(2x) + \cos(3x)] = 0$

   Vậy $\cos x = 0$ hoặc $\cos (2x) + \cos(3x) = 0$

   TH1: $\cos x = 0$

   Suy ra $x = \dfrac{\pi}{2} + k\pi$

   TH2: $\cos(2x) = -\cos(3x)$

   Suy ra 

   $\cos(2x) = \cos (\pi – 3x)$

   $<-> 2x = \pi – 3x + 2k\pi$ hoặc $2x = 3x – \pi + 2k\pi$

   $<-> x = \dfrac{\pi}{5} + \dfrac{2k\pi}{5}$ hoặc $x = \pi + 2k\pi$.

   Trả lời