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Question Number 111513 by john santu last updated on 04/Sep/20

$$\sqrt[3]{\cos {}^{2}x}+\sqrt[3]{\sin {}^{2}x}=\sqrt[3]{2}$$$$find\cos {}^2\left(2x\right)?$$

Answered by bemath last updated on 04/Sep/20

$${(\sqrt[3]{\cos {}^2\mathrm{x}}+\sqrt[3]{\sin {}^2\mathrm{x}})}^3=2$$$${(\mathrm{a}+\mathrm{b})}^3={\mathrm{a}}^3+{\mathrm{b}}^3+3\mathrm{ab}(\mathrm{a}+\mathrm{b})$$$$\Rightarrow 1+3\sqrt[3]{\sin {}^2\mathrm{x}\cos {}^2\mathrm{x}}(\sqrt[3]{\sin {}^2\mathrm{x}}+\sqrt[3]{\cos {}^2\mathrm{x}})=2$$$$\Rightarrow 3.\sqrt[3]{2}\sqrt[3]{\frac{1}{4}\sin {}^2\left(2\mathrm{x}\right)}=1$$$$\Rightarrow 54\left(\frac{1}{4}\sin {}^2\left(2\mathrm{x}\right)\right)=1$$$$\Rightarrow \sin {}^2\left(2\mathrm{x}\right)=\frac{4}{54}\to \cos {}^2\left(2\mathrm{x}\right)=\frac{54-4}{54}$$$$\to \cos {}^2\left(2\mathrm{x}\right)=\frac{50}{54}=\frac{25}{27}$$

Answered by som(math1967) last updated on 04/Sep/20

$${\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}+3^3\sqrt{{\cos }^2{\mathrm{xsin}}^2\mathrm{x}}\left\{{}^3\sqrt{2}\right\}=2$$$$27\times {2\sin }^2{\mathrm{xcos}}^2\mathrm{x}=1$$$${\left(2\mathrm{sinxcosx}\right)}^2=\frac{2}{27}$$$${\sin }^{2}2\mathrm{x}=\frac{2}{27}$$$${\cos }^{2}2\mathrm{x}=1-\frac{2}{27}=\frac{25}{27}$$

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