Question

2cos²x+senx=1
Es una ecuación para 0≤×<π

Answer 1

2cos²x+senx = 1 0≤x<π

2(1-sen²x)+senx = 1

2 - 2sen²x + senx = 1

2sen²x - senx - 1 = 0

(senx - 1)(2senx + 1) = 0

senx = 1 ó senx = -1/2

x = π/2 ó x = -π/6