Matematika

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Cos 3x - cos 2x + cos x = 0

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  • cosx−cos2x+cos3x=0cos⁡x−cos⁡2x+cos⁡3x=0(cosx+cos3x)−cos2x=0(cos⁡x+cos⁡3x)−cos⁡2x=02cos(x+3x2)cos(x−3x2)−cos2x=02cos⁡(x+3x2)cos⁡(x−3x2)−cos⁡2x=0

    2cos2xcosx−cos2x=02cos⁡2xcos⁡x−cos⁡2x=0cos2x(2cosx−1)=0cos⁡2x(2cos⁡x−1)=0Now, solving for xx as followscos2x=0⟹2x=(2k+1)π2cos⁡2x=0⟹2x=(2k+1)π2x=(2k+1)π4x=(2k+1)π4x=90∘k+45∘x=90∘k+45∘or2cosx−1=02cos⁡x−1=0cosx=12=cosπ3cos⁡x=12=cos⁡π3x=2kπ±π3x=2kπ±π3x=360∘60∘

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