Question-120882

Question Number 120882 by shaker last updated on 03/Nov/20

Answered by $@y@m last updated on 03/Nov/20

cos^3 x+cos x=1−cos^2 x cos x(cos^2 x+1)=sin^2 x cos x(2−sin^2 x)=sin^2 x cos^2 x(2−sin^2 x)^2 =sin^4 x (1−sin^2 x)(4+sin^4 x−4sin^2 x)=sin^4 x 4+sin^4 x−4sin^2 x−4sin^2 x−sin^6 x+4sin^4 x=sin^4 x sin^6 x−4sin^4 x+8sin^2 x−4=0 sin^6 x−4sin^4 x+8sin^2 x+3=7

$$\mathrm{cos}^{\mathrm{3}} {x}+\mathrm{cos}\:{x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x} \\ $$$$\mathrm{cos}\:{x}\left(\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{1}\right)=\mathrm{sin}^{\mathrm{2}} {x} \\ $$$$\mathrm{cos}\:{x}\left(\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)=\mathrm{sin}^{\mathrm{2}} {x} \\ $$$$\mathrm{cos}^{\mathrm{2}} {x}\left(\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} =\mathrm{sin}^{\mathrm{4}} {x} \\ $$$$\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)\left(\mathrm{4}+\mathrm{sin}\:^{\mathrm{4}} {x}−\mathrm{4sin}\:^{\mathrm{2}} {x}\right)=\mathrm{sin}^{\mathrm{4}} {x} \\ $$$$\mathrm{4}+\mathrm{sin}\:^{\mathrm{4}} {x}−\mathrm{4sin}\:^{\mathrm{2}} {x}−\mathrm{4sin}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{4sin}\:^{\mathrm{4}} {x}=\mathrm{sin}\:^{\mathrm{4}} {x} \\ $$$$\mathrm{sin}\:^{\mathrm{6}} {x}−\mathrm{4sin}\:^{\mathrm{4}} {x}+\mathrm{8sin}\:^{\mathrm{2}} {x}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{sin}\:^{\mathrm{6}} {x}−\mathrm{4sin}\:^{\mathrm{4}} {x}+\mathrm{8sin}\:^{\mathrm{2}} {x}+\mathrm{3}=\mathrm{7} \\ $$

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Question-120882

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