Question

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фото внизу

Answer 1

$frac{sin x}{sin x-cos x}=sqrt3left(frac{cos x}{sin x+cos x}+tg2xright)\sqrt3left(frac{cos x}{sin x+cos x}+tg2xright)=sqrt3left(frac{cos x}{sin x+cos x}+frac{2}{ctgx-tgx}right)=\=sqrt3left(frac{cos x}{sin x+cos x}+frac{2}{frac{cos x}{sin x}-frac{sin x}{cos x}}right)=sqrt3left(frac{cos x}{sin x+cos x}+frac{2}{frac{cos^2x-sin^2x}{sin xcos x}}right)=$
$=sqrt3left(frac{cos x}{sin x+cos x}+frac{2sin xcos x}{{cos^2x-sin^2x}{}}right)=\=sqrt3left(frac{cos x}{sin x+cos x}+frac{2sin xcos x}{{(cos x-sin x)(cos x+sin x)}{}}right)=\=sqrt3left(frac{cos x(cos x-sin x)+2sin xcos x}{(cos x-sin x)(cos x+sin x)}right)=sqrt3left(frac{cos^2x-cos xsin x+2sin xcos x}{(cos x-sin x)(cos x+sin x)}right)=\=sqrt3left(frac{cos x(cos x+sin x)}{(cos x-sin x)(cos x+sin x)}right)=sqrt3cdotfrac{cos x}{cos x-sin x}$
$frac{sin x}{sin x-cos x}=sqrt3frac{cos x}{cos x-sin x}\frac{sin x}{sin x-cos x}=-sqrt3frac{cos x}{sin x-cos x}\frac{sin x}{cos x}=-sqrt3\tgx=-sqrt3Rightarrow x=frac{2pi}3+pi n,;ninmathbb{Z}$