Question

тригонометрия
решите пожалуйста ​

Answer 1

$\dfrac{2sinx-cosx}{sinx+cosx}=\dfrac{2}{3}\ \ \ \Rightarrow \ \ \ 3\, (2sinx-cosx)=2\, (sinx+cosx)\\\\\\6sinx-3cosx=2sinx+2cosx\ \ ,\ \ \ 4sinx=5cosx\ \ \to \ \ \dfrac{sinx}{cosx}=\dfrac{5}{4}\ ,\\\\tgx=\dfrac{5}{4}\ \ ,\ \ \ \ sinx=\dfrac{5}{4}\, cosx\\\\\\\boxed{\ 1+tg^2x=\dfrac{1}{cos^2x}\ }\ \ \ \Rightarrow \ \ \ \ cos^2x=\dfrac{1}{1+tg^2x}=\dfrac{1}{1+(\frac{5}{4})^2}=\dfrac{16}{16+25}=\dfrac{16}{41}$

$sin^2x-cos^2x=\Big(\dfrac{5}{4}\, cosx\Big)^2-cos^2x=\dfrac{25}{16}\, cos^2x-cos^2x=\dfrac{9}{16}\, cos^2x=\\\\\\=\dfrac{9}{16}\cdot \dfrac{16}{41}=\dfrac{9}{41}\\\\\\ili\ \ \ sin^2x-cos^2x=(1-cos^2x)-cos^2x=1-2cos^2x=1-2\cdot \dfrac{16}{41}=\\\\\\=\dfrac{41-32}{41}=\dfrac{9}{41}$