Question

Решите,пожалуйста,тригонометрическое уравнение.
sin^2x-3cosx=3
cosx+sin2x=0

Answer 1

sin²x - 3cosx - 3 = 0

1 - cos²x - 3cosx - 3 = 0

cos²x + 3cosx +2 = 0

cosx = t

t² + 3t + 2 = 0

D = 9 - 8 = 1

t1 = 0.5( - 3 - 1) = -2 не подходит, так как t = соsx ≤1

t2 = 0.5( -3 + 1) = -1

cosx = -1

x = π + 2πk k∈Z

cosx + sin2x = 0

cosx + 2 cosx · sinx = 0

cosx (1 + 2sinx) = 0

1) cosx = 0

x1 = π/2 + πk k∈Z

2) 1 + 2sinx = 0

2sinx = -1

sinx = -1/2

x2 = -π/2 + 2πk k∈Z

x3 = -5π/6 + 2πk k∈z