Erin Lozano
2022-06-24
Answered

Solve the following equation: $\begin{array}{r}{\mathrm{sin}}^{14}x+{\mathrm{cos}}^{14}x=\frac{169}{64}{\mathrm{cos}}^{6}2x\end{array}$

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Paxton James

Answered 2022-06-25
Author has **25** answers

$\begin{array}{rl}2{\mathrm{sin}}^{2}x& =1-\mathrm{cos}2x\\ 2{\mathrm{cos}}^{2}x& =1+\mathrm{cos}2x\\ 128({\mathrm{sin}}^{14}x+{\mathrm{cos}}^{14}x)& =(1-\mathrm{cos}2x{)}^{7}+(1+\mathrm{cos}2x{)}^{7}\\ & =2(1+21{\mathrm{cos}}^{2}2x+35{\mathrm{cos}}^{4}2x+7{\mathrm{cos}}^{6}2x)\\ 169{\mathrm{cos}}^{6}2x& =1+21{\mathrm{cos}}^{2}2x+35{\mathrm{cos}}^{4}2x+7{\mathrm{cos}}^{6}2x\\ 0& =162{\mathrm{cos}}^{6}2x-35{\mathrm{cos}}^{4}2x-21{\mathrm{cos}}^{2}2x-1\\ 0& =(2{\mathrm{cos}}^{2}2x-1)(81{\mathrm{cos}}^{4}2x+23{\mathrm{cos}}^{2}2x+1)\\ {\mathrm{cos}}^{2}2x& =\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{-23\pm \sqrt{{23}^{2}-4(81)}}{2(81)}<0\end{array}$

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let

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