cos-2-x-1-3-sin-2-x-1-3-2-1-3-find-cos-2-2x-

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}

f i n d \cos^{2} (2 x) ?

Answered by bemath last updated on 04/Sep/20

{(\sqrt[3]{\cos^{2} x} + \sqrt[3]{\sin^{2} x})}^{3} = 2

{(a + b)}^{3} = a^{3} + b^{3} + 3 ab (a + b)

\Rightarrow 1 + 3 \sqrt[3]{\sin^{2} x \cos^{2} x} (\sqrt[3]{\sin^{2} x} + \sqrt[3]{\cos^{2} x}) = 2

\Rightarrow 3 . \sqrt[3]{2} \sqrt[3]{\frac{1}{4} \sin^{2} (2 x)} = 1

\Rightarrow 54 (\frac{1}{4} \sin^{2} (2 x)) = 1

\Rightarrow \sin^{2} (2 x) = \frac{4}{54} \to \cos^{2} (2 x) = \frac{54 - 4}{54}

\to \cos^{2} (2 x) = \frac{50}{54} = \frac{25}{27}

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

\cos^{2} x + \sin^{2} x + 3^{3} \sqrt{\cos^{2} {xsin}^{2} x} {^{3} \sqrt{2}} = 2

27 \times {2 \sin}^{2} {xcos}^{2} x = 1

{(2 sinxcosx)}^{2} = \frac{2}{27}

\sin^{2} 2 x = \frac{2}{27}

\cos^{2} 2 x = 1 - \frac{2}{27} = \frac{25}{27}