Question

решите уравнение: 2 sin в квадрате x +5 cos x +1 = 0

Answer 1

 

$<var>2sin^2x + 5cosx + 1 = 0\\\\ 2(1 - cos^2x) + 5cosx + 1 = 0\\\\ 2 - 2cos^2x + 5cosx + 1 = 0\\\\ - 2cos^2x + 5cosx + 3 = 0\\\\ cosx = t\\\\ -2t^2+5t+3 = 0\\\\ t_1*t_2 = -\frac{3}{2}\\\\ t_1 + t_2 = \frac{5}{2}\\\\ t_1 = -\frac{1}{2}\\\\ t_2 = 3\\\\ cosx = -\frac{1}{2}\\\\ x = \frac{2\pi}{3} + 2\pi n\\\\ x = -\frac{2\pi}{3} + 2\pi n</var>$

 

Answer 2

2соs^2х+1+5соsх+1=0

2соs^2х+5сosx+2=0

Пусть соsх=t

2t^2+5t+2=0

D=25-16=9=3

t1=-2

t2=2/4

х1=4

х2=2,35