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resoudre-dans-pi-pi-cosx-cos2x-cos3x-0-




Question Number 198483 by Guillaume last updated on 20/Oct/23
resoudre dans ]−π;π]  cosx+cos2x+cos3x=0
resoudredans]π;π]cosx+cos2x+cos3x=0
Answered by Frix last updated on 21/Oct/23
c=cos x  c+2c^2 −1+4c^3 −3c=0  c^3 +(c^2 /2)−(c/2)−(1/4)=0  (c+((√2)/2))(c+(1/2))(c−((√2)/2))=0  x=±((3π)/4)∨x=±((2π)/3)∨x=±(π/4)
c=cosxc+2c21+4c33c=0c3+c22c214=0(c+22)(c+12)(c22)=0x=±3π4x=±2π3x=±π4
Answered by mr W last updated on 21/Oct/23
cos x+2 cos^2  x−1+4 cos^3  x−3 cos x=0  4 cos^3  x+2 cos^2  x−2 cos x−1=0  let u=cos x  4u^3 +2u^2 −2u−1=0  (2u+1)(2u^2 −1)=0  ⇒u=−(1/2) ⇒x=±120°  ⇒u=((√2)/2) ⇒x=±45°  ⇒u=−((√2)/2) ⇒x=±135°
cosx+2cos2x1+4cos3x3cosx=04cos3x+2cos2x2cosx1=0letu=cosx4u3+2u22u1=0(2u+1)(2u21)=0u=12x=±120°u=22x=±45°u=22x=±135°
Answered by cortano12 last updated on 21/Oct/23
 cos 2x +(cos 3x+cos x)=0   cos 2x+2cos 2x cos x = 0   cos 2x(1+2cos x)=0     { ((cos 2x=0⇒x=± (π/4))),((cos x=−(1/2)⇒x=±((2π)/3))),((cos 2x=cos ((3π)/2)⇒x=±((3π)/4))) :}
cos2x+(cos3x+cosx)=0cos2x+2cos2xcosx=0cos2x(1+2cosx)=0{cos2x=0x=±π4cosx=12x=±2π3cos2x=cos3π2x=±3π4

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