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Question Number 185134 by emmanuelson123 last updated on 17/Jan/23

Answered by som(math1967) last updated on 17/Jan/23

$$\frac{{sin}^{4}x}{5}+\frac{{(1-{sin}^{2}x)}^2}{7}=\frac{1}{12}$$$$\Rightarrow 7{sin}^{4}x+5-10{sin}^{2}x+5{sin}^{4}x=\frac{35}{12}$$$$\Rightarrow 144{sin}^{4}x-120{sin}^{2}x+60=35$$$$\Rightarrow {(12{sin}^{2}x-5)}^2=0$$$$\Rightarrow {sin}^{2}x=\frac{5}{12}$$$$\Rightarrow 2{sin}^{2}x=\frac{5}{6}$$$$\Rightarrow 1-cos2x=\frac{5}{6}$$$$\Rightarrow cos2x=\frac{1}{6}$$$$\Rightarrow {cos}^{2}2x=\frac{1}{36}$$$$\Rightarrow {sin}^{2}2x=1-\frac{1}{36}=\frac{35}{36}$$$$\therefore \frac{{sin}^{2}2x}{5}+\frac{{cos}^{2}2x}{7}$$$$=\frac{35}{5\times 36}+\frac{1}{36\times 7}$$$$=\frac{7}{36}+\frac{1}{36\times 7}=\frac{1}{36}\times \frac{50}{7}=\frac{25}{126}$$

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