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Question Number 185257 by Kalebwizeman last updated on 19/Jan/23

Answered by SEKRET last updated on 19/Jan/23

$$\frac{25}{126}$$

Answered by SEKRET last updated on 19/Jan/23

$$Q185134$$

Commented by Kalebwizeman last updated on 19/Jan/23

$$ohthanks$$

Answered by a.lgnaoui last updated on 19/Jan/23

$$Methode1$$$${(\sin {}^{2}x+\cos {}^{2}x)}^2={\sin }^{4}x+\cos {}^{4}x+2\sin x^2\cos x^2$$$$1={\sin }^{4}x+{\cos }^{4}x+\frac{{\sin }^{2}2x}{2}\left(1\right)$$$${\cos }^{2}x-{\sin }^{2}x=\cos 2x={2\cos }^{2}x-1$$$${\cos }^{2}x=\frac{\cos 2x+1}{2}\left(2\right)$$$$\left(1\right)\iff 1={\sin }^{4}x+{\left(\frac{\cos 2x+1}{2}\right)}^2+\frac{{\sin }^{2}2x}{2}$$$$\frac{{(\cos 2x+1)}^2+{2\sin }^{2}2x}{4}=1-{\sin }^{4}x\left(i\right)$$$$dapres\frac{{\sin }^{4}x}{5}+\frac{{\cos }^{4}x}{7}=\frac{1}{12}$$$${\sin }^{4}x=\frac{5}{12}-\frac{5\cos x^4}{7}$$$$$$$$\frac{5\cos {}^{4}x}{7}+\frac{7}{12}=\frac{{(\cos 2x+1)}^2+{2\sin }^{2}2x}{4}$$$$\frac{{(\cos 2\mathrm{x}+1)}^2+{2\sin }^{2}2\mathrm{x}}{4}-\frac{5{(\cos 2x+1)}^2}{28}=\frac{7}{12}$$$$[\frac{{(\cos 2\mathrm{x}+1)}^2}{4}-\frac{5{(\cos 2\mathrm{x}+1)}^2}{28}]+\frac{2(1-\cos {}^{2}2\mathrm{x})}{4}=\frac{7}{12}$$$$$$$${\cos }^{2}2x-\frac{\cos 2x}{3}+\frac{1}{36}=0$$$$△=0$$$$\cos 2x=\frac{1}{6}\Rightarrow \sin 2x=\frac{\sqrt{35}}{6}$$$$donc$$$$\frac{\sin {}^{2}2x}{5}+\frac{\cos 2x{}^2}{7}=\frac{\frac{35}{36}}{5}+\frac{\frac{1}{36}}{7}$$$$$$$$[\frac{{\sin }^{2}2x}{5}+\frac{{\cos }^{2}2x}{7}=\frac{25}{126}$$$$$$

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