Question

sabendo que cosx - senx =√6/3 entao o valor de sen (2x) é?

Answer 1

$\cos x-\mathrm{sen\,}x=\frac{\sqrt{6}}{3}$


Elevando os dois lados ao quadrado, temos

$(\cos x-\mathrm{sen\,}x)^{2}=\left(\frac{\sqrt{6}}{3} \right )^{2}\\ \\ \cos^{2}x-2\,\mathrm{sen\,}x\cos x+\mathrm{sen^{2}\,}x=\frac{6}{9}\\ \\ (\cos^{2}x+\mathrm{sen^{2}\,}x)-2\,\mathrm{sen\,}x\cos x=\frac{2}{3}\\ \\ 1-2\,\mathrm{sen\,}x\cos x=\frac{2}{3}\\ \\ 2\,\mathrm{sen\,}x\cos x=1-\frac{2}{3}\\ \\ 2\,\mathrm{sen\,}x\cos x=\frac{3-2}{3}\\ \\ 2\,\mathrm{sen\,}x\cos x=\frac{1}{3}\\ \\ \boxed{\begin{array}{c}\mathrm{sen}(2x)=\frac{1}{3} \end{array}}$