Question

Cos^2x+2*sinx*cosx-3*sin^2x=0 решить

Answer 1

Cos²x + 2 sinx cosx - 3 sin²x = 0 | : cos²x
1 + 2 tgx - 3tg²x = 0 | * (-1)
3tg²x - 2tgx -1 = 0
Пусть tgx = t, тогда
3t² - 2t -1 = 0
D = b² - 4ac = 4 + 12 = 16
$x_{1} = \frac{2+4}{6}$
$x_{1} = 1$
$x_{2} = \frac{2 - 4}{6}$
$x_{2} = \frac{-1}{3}$
Значит 1) tgx = 1
X = $\frac{ \pi }{4}$ + $\pi n$
2) tgx = - $\frac{1}{3}$
x = - arctgx $\frac{1}{3}$ + $\pi n$
n ∈ Z