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Question-185134




Question Number 185134 by emmanuelson123 last updated on 17/Jan/23
Answered by som(math1967) last updated on 17/Jan/23
 ((sin^4 x)/5) +(((1−sin^2 x)^2 )/7)=(1/(12))  ⇒7sin^4 x+5−10sin^2 x+5sin^4 x=((35)/(12))  ⇒144sin^4 x−120sin^2 x+60=35  ⇒(12sin^2 x−5)^2 =0  ⇒ sin^2 x=(5/(12))  ⇒2sin^2 x=(5/6)  ⇒1−cos2x=(5/6)  ⇒cos2x=(1/6)  ⇒cos^2 2x=(1/(36))  ⇒sin^2 2x=1−(1/(36))=((35)/(36))  ∴((sin^2 2x)/5) +((cos^2 2x)/7)  =((35)/(5×36)) +(1/(36×7))  =(7/(36)) +(1/(36×7))=(1/(36))×((50)/7)=((25)/(126))
$$\:\frac{{sin}^{\mathrm{4}} {x}}{\mathrm{5}}\:+\frac{\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{7}{sin}^{\mathrm{4}} {x}+\mathrm{5}−\mathrm{10}{sin}^{\mathrm{2}} {x}+\mathrm{5}{sin}^{\mathrm{4}} {x}=\frac{\mathrm{35}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{144}{sin}^{\mathrm{4}} {x}−\mathrm{120}{sin}^{\mathrm{2}} {x}+\mathrm{60}=\mathrm{35} \\ $$$$\Rightarrow\left(\mathrm{12}{sin}^{\mathrm{2}} {x}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{sin}^{\mathrm{2}} {x}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{2}{sin}^{\mathrm{2}} {x}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{1}−{cos}\mathrm{2}{x}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\Rightarrow{cos}\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} \mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{36}} \\ $$$$\Rightarrow{sin}^{\mathrm{2}} \mathrm{2}{x}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{36}}=\frac{\mathrm{35}}{\mathrm{36}} \\ $$$$\therefore\frac{{sin}^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{5}}\:+\frac{{cos}^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{7}} \\ $$$$=\frac{\mathrm{35}}{\mathrm{5}×\mathrm{36}}\:+\frac{\mathrm{1}}{\mathrm{36}×\mathrm{7}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{36}}\:+\frac{\mathrm{1}}{\mathrm{36}×\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{36}}×\frac{\mathrm{50}}{\mathrm{7}}=\frac{\mathrm{25}}{\mathrm{126}} \\ $$

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