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Question Number 198483 by Guillaume last updated on 20/Oct/23
resoudre dans ]−π;π] cosx+cos2x+cos3x=0
$$\left.{r}\left.{esoudre}\:{dans}\:\right]−\pi;\pi\right] \\ $$$${cosx}+{cos}\mathrm{2}{x}+{cos}\mathrm{3}{x}=\mathrm{0}\: \\ $$
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)
$${c}=\mathrm{cos}\:{x} \\ $$$${c}+\mathrm{2}{c}^{\mathrm{2}} −\mathrm{1}+\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c}=\mathrm{0} \\ $$$${c}^{\mathrm{3}} +\frac{{c}^{\mathrm{2}} }{\mathrm{2}}−\frac{{c}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({c}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({c}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${x}=\pm\frac{\mathrm{3}\pi}{\mathrm{4}}\vee{x}=\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\vee{x}=\pm\frac{\pi}{\mathrm{4}} \\ $$
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°
$$\mathrm{cos}\:{x}+\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{1}+\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:{x}−\mathrm{3}\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:{x}+\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{2}\:\mathrm{cos}\:{x}−\mathrm{1}=\mathrm{0} \\ $$$${let}\:{u}=\mathrm{cos}\:{x} \\ $$$$\mathrm{4}{u}^{\mathrm{3}} +\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{u}−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}{u}+\mathrm{1}\right)\left(\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{x}=\pm\mathrm{120}° \\ $$$$\Rightarrow{u}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow{x}=\pm\mathrm{45}° \\ $$$$\Rightarrow{u}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow{x}=\pm\mathrm{135}° \\ $$
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))) :}
$$\:\mathrm{cos}\:\mathrm{2x}\:+\left(\mathrm{cos}\:\mathrm{3x}+\mathrm{cos}\:\mathrm{x}\right)=\mathrm{0} \\ $$$$\:\mathrm{cos}\:\mathrm{2x}+\mathrm{2cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{0} \\ $$$$\:\mathrm{cos}\:\mathrm{2x}\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{x}\right)=\mathrm{0} \\ $$$$\:\:\begin{cases}{\mathrm{cos}\:\mathrm{2x}=\mathrm{0}\Rightarrow\mathrm{x}=\pm\:\frac{\pi}{\mathrm{4}}}\\{\mathrm{cos}\:\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}=\pm\frac{\mathrm{2}\pi}{\mathrm{3}}}\\{\mathrm{cos}\:\mathrm{2x}=\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\Rightarrow\mathrm{x}=\pm\frac{\mathrm{3}\pi}{\mathrm{4}}}\end{cases} \\ $$

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