Question

Необходимо решить уравнение:     (1/tg^2 x) - (1/sin x) - 1 = 0

Answer 1

$frac1{tg^2x} -frac1{sin x} - 1 = 0\frac1{frac{sin^2x}{cos^2x}}-frac1{sin x}-1=0\frac{cos^2x}{sin^2x}-frac1{sin x}-1=0\frac{1-sin^2x}{sin^2x}-frac1{sin x}-1=0\frac1{sin^2x}-1-frac1{sin x}-1=0\frac1{sin^2x}-frac1{sin x}-2=0\sin xneq0Rightarrow xneqpi n,;ninmathbb{Z}\frac1{sin x}=t,;frac1{sin^2x}=t^2,;tneq0\t^2-t-2=0\D=1+4cdot2=9\t_1=-1,;t_2=2\frac1{sin x}=-1Rightarrowsin x=-1Rightarrow x=pi+2pi n,;ninmathbb{Z}$
$frac1{sin x}=2Rightarrowsin x=frac12Rightarrow x=fracpi6+2pi n,;ninmathbb{Z}$