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If $\sin x+\sin ^{2} x=1$, then $\cos ^{8} x+2 \cos ^{6} x+\cos ^{4} x$ is
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Given $\sin x+\sin ^{2} x=1 \Rightarrow \sin x=1-\sin ^{2} x$
$\begin{aligned} \therefore \sin x=\cos ^{2} x \Rightarrow \sin ^{2} x &=\cos ^{4} x \\ \cos ^{8} x+2 \cos ^{6} x+\cos ^{4} x &=\left(\cos ^{4} x+\cos ^{2} x\right)^{2} \\ &=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}=1 \end{aligned}$
$\begin{aligned} \therefore \sin x=\cos ^{2} x \Rightarrow \sin ^{2} x &=\cos ^{4} x \\ \cos ^{8} x+2 \cos ^{6} x+\cos ^{4} x &=\left(\cos ^{4} x+\cos ^{2} x\right)^{2} \\ &=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}=1 \end{aligned}$
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