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Question Number 77041 by mathocean1 last updated on 02/Jan/20

solve in [0;π] sinx−sin^3 x=1−cos2x

$$\mathrm{solve}\:\mathrm{in}\:\left[\mathrm{0};\pi\right] \\ $$$$\mathrm{sin}{x}−\mathrm{sin}^{\mathrm{3}} {x}=\mathrm{1}−\mathrm{cos2}{x} \\ $$

Answered by mind is power last updated on 02/Jan/20

⇔ 1 sin(x)(1−sin^2 (x))=2sin^2 (x) ⇔sin(x)(sin^2 (x)+2sin(x)−1)=0 easy Know

$$\Leftrightarrow \\ $$$$\mathrm{1} \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)=\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\Leftrightarrow\mathrm{sin}\left(\mathrm{x}\right)\left(\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2sin}\left(\mathrm{x}\right)−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{easy}\:\mathrm{Know} \\ $$

Answered by mr W last updated on 02/Jan/20

sinx−sin^3 x=1−cos2x sinx−sin^3 x=1−1+2 sin^2 x sinx−sin^3 x−2 sin^2 x=0 sinx(1−2 sin x−sin^2 x)=0 ⇒sin x=0 ⇒x=0, π ⇒sin x=(√2)−1 ⇒x=sin^(−1) ((√2)−1), π−sin^(−1) ((√2)−1)

$$\mathrm{sin}{x}−\mathrm{sin}^{\mathrm{3}} {x}=\mathrm{1}−\mathrm{cos2}{x} \\ $$$$\mathrm{sin}{x}−\mathrm{sin}^{\mathrm{3}} {x}=\mathrm{1}−\mathrm{1}+\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{sin}{x}−\mathrm{sin}^{\mathrm{3}} {x}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}=\mathrm{0} \\ $$$$\mathrm{sin}{x}\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:{x}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\mathrm{0}\:\Rightarrow{x}=\mathrm{0},\:\pi \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\sqrt{\mathrm{2}}−\mathrm{1}\:\Rightarrow{x}=\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right),\:\pi−\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$

Commented by mr W last updated on 02/Jan/20

Dear sir! Happy new year to you too!

$${Dear}\:{sir}!\:\:{Happy}\:{new}\:{year}\:{to}\:{you}\:{too}! \\ $$

Commented by mind is power last updated on 02/Jan/20

Hello Mr How are You sir ?happy new year

$$\mathrm{Hello}\:\mathrm{Mr}\:\mathrm{How}\:\mathrm{are}\:\mathrm{You}\:\mathrm{sir}\:?\mathrm{happy}\:\mathrm{new}\:\mathrm{year} \\ $$

Commented by mathocean1 last updated on 03/Jan/20

thank you sirs

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sirs} \\ $$

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