Question

Resolvendo a equação 2sen²x = 1 + cos x no intervalo 0 ≤ X < 2π, a soma das raízes vale:
a) 2π
b) 3π
c) 4π
d)3π/2
e)5π/2

Answer 1

Boa noite

2sen²(x) = 1 + cos(x)

sen²(x) + cos²(x) = 1
sen²(x) = 1 - cos²(x)

2 - 2cos²(x) = 1 + cos(x)

2cos²(x) + cos(x) - 1 = 0

y = cos(x)

2y² + y - 1 = 0

delta
d² = 1 + 8 = 9
d = 3

y1 = (-1 + 3)/4 = 1/2
y2 = (-1 - 3)/4 = -1

cos(x) = 1/2
cos(x) = -1, x = π

x = π/3 
x = π/6
x = π

soma
S = π + π/3 + 5π/6 = 3π/3 + π/3 + 5π/3= 9π/3 = 3π (B)