Question

Очень Срочно!!!Помогите, пожалуйста, с решением:

1)Ctg α = - 5/12; 3π/2<α<2π;

Sin=?

2) 25cos² α - 5cos α - 12 = 0

π<α<3π/2; ctg α = ?

Answer 1

$1)1+Ctg^{2}\alpha =\frac{1}{Sin^{2}\alpha} \\\\Sin^{2}\alpha=\frac{1}{1+Ctg^{2}\alpha}=\frac{1}{1+(-\frac{5}{12})^{2}}=\frac{1}{1+\frac{25}{144}}=\frac{1}{\frac{169}{144}}=\frac{144}{169}\\\\Sin\alpha =-\sqrt{\frac{144}{169} }=-\frac{12}{13}$

2) 25Cos²α - 5Cosα - 12 = 0

Обозначим Cosα = m , - 1 ≤ m ≤ 1

25m² - 5m - 12 = 0

D = (- 5)² - 4 * 25 * (- 12) = 25 + 1200 = 1225 = 35²

$m_{1}=\frac{5+35}{50}=\frac{4}{5}=0,8\\\\m_{2}=\frac{5-35}{50}=-0,6\\\\Cos\alpha=0,8\\\\\alpha=+-arcCos0,8+2\pi n,nez\\\\Cos\alpha =-0,6\\\\\alpha=+-arcCos(-0,6)+2\pi n,nez$