Question Number 185257 by Kalebwizeman last updated on 19/Jan/23
Answered by SEKRET last updated on 19/Jan/23
$$\frac{\mathrm{25}}{\mathrm{126}} \\ $$
Answered by SEKRET last updated on 19/Jan/23
$${Q}\mathrm{185134} \\ $$
Commented by Kalebwizeman last updated on 19/Jan/23
$${oh}\:{thanks} \\ $$
Answered by a.lgnaoui last updated on 19/Jan/23
![Methode 1 (sin^2 x+cos^2 x)^2 =sin^4 x+cos^4 x+2sinx^2 cosx^2 1=sin^4 x+cos^4 x+((sin^2 2x )/2) (1) cos^2 x−sin^2 x=cos 2x=2cos^2 x−1 cos^2 x=((cos 2x+1)/2) (2) (1)⇔1=sin^4 x+(((cos2x+1 )/2))^2 +((sin^2 2x )/2) (((cos2x+1)^2 +2sin^2 2x )/4)=1−sin^4 x(i) d apres ((sin^4 x)/5) +((cos^4 x )/7)=(1/(12)) sin^4 x=(5/(12)) −((5cosx^4 )/7) ((5cos^4 x)/7)+(7/(12))=(((cos2x+1)^2 +2sin^2 2x )/4) (((cos2x+1)^2 +2sin^2 2x )/4)−((5(cos 2x+1)^2 )/(28))=(7/(12)) [(((cos 2x+1)^2 )/4)−((5(cos 2x+1)^2 )/(28))]+((2(1−cos^2 2x) )/4)=(7/(12)) cos^2 2x−((cos 2x)/3) +(1/(36))=0 △=0 cos 2x=(1/6) ⇒sin 2x=((√(35))/6) donc ((sin^2 2x)/5)+((cos2x^2 )/7) =(((35)/(36))/5)+((1/(36))/7) [ ((sin^2 2x )/5)+((cos^2 2x )/7) =((25)/(126))](Q185275.png)
$${Methode}\:\mathrm{1} \\ $$$$\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} =\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}+\mathrm{2sin}{x}^{\mathrm{2}} \mathrm{cos}{x}^{\mathrm{2}} \\ $$$$\mathrm{1}=\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{cos}^{\mathrm{4}} {x}+\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}\:}{\mathrm{2}}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2cos}^{\mathrm{2}} {x}−\mathrm{1} \\ $$$$\mathrm{cos}^{\mathrm{2}} {x}=\frac{\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\mathrm{1}=\mathrm{sin}^{\mathrm{4}} {x}+\left(\frac{\mathrm{cos2}{x}+\mathrm{1}\:}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}\:}{\mathrm{2}}\: \\ $$$$\frac{\left(\mathrm{cos2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2sin}^{\mathrm{2}} \mathrm{2}{x}\:\:}{\mathrm{4}}=\mathrm{1}−\mathrm{sin}^{\mathrm{4}} {x}\left({i}\right) \\ $$$${d}\:{apres}\:\frac{\mathrm{sin}^{\mathrm{4}} {x}}{\mathrm{5}}\:+\frac{\mathrm{cos}^{\mathrm{4}} {x}\:}{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{sin}^{\mathrm{4}} {x}=\frac{\mathrm{5}}{\mathrm{12}}\:−\frac{\mathrm{5cos}{x}^{\mathrm{4}} \:}{\mathrm{7}} \\ $$$$ \\ $$$$\frac{\mathrm{5cos}\:^{\mathrm{4}} {x}}{\mathrm{7}}+\frac{\mathrm{7}}{\mathrm{12}}=\frac{\left(\mathrm{cos2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2sin}^{\mathrm{2}} \mathrm{2}{x}\:}{\mathrm{4}} \\ $$$$\frac{\left(\mathrm{cos2x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2sin}^{\mathrm{2}} \mathrm{2x}\:\:}{\mathrm{4}}−\frac{\mathrm{5}\left(\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{28}}=\frac{\mathrm{7}}{\mathrm{12}} \\ $$$$\left[\frac{\left(\mathrm{cos}\:\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{5}\left(\mathrm{cos}\:\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{28}}\right]+\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2x}\right)\:}{\mathrm{4}}=\frac{\mathrm{7}}{\mathrm{12}} \\ $$$$ \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2}{x}−\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{36}}=\mathrm{0} \\ $$$$\:\:\:\:\bigtriangleup=\mathrm{0} \\ $$$$\:\mathrm{cos}\:\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\Rightarrow\mathrm{sin}\:\mathrm{2}{x}=\frac{\sqrt{\mathrm{35}}}{\mathrm{6}} \\ $$$${donc} \\ $$$$\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{5}}+\frac{\mathrm{cos2}{x}\:^{\mathrm{2}} }{\mathrm{7}}\:=\frac{\frac{\mathrm{35}}{\mathrm{36}}}{\mathrm{5}}+\frac{\frac{\mathrm{1}}{\mathrm{36}}}{\mathrm{7}} \\ $$$$ \\ $$$$\:\:\:\:\:\left[\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}\:}{\mathrm{5}}+\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{2}{x}\:}{\mathrm{7}}\:=\frac{\mathrm{25}}{\mathrm{126}}\:\right. \\ $$$$\:\:\:\:\:\:\:\: \\ $$