Solve the following equation in the interval [0 , 2π).

Question

sin²(x) = 3 - 2sin(x)

a.

x = π/3

b.

x = π

c.

x = 2π

d.

x = π/2

Answer 1

let y = sin(x)

y²+2y-3=0

(y+3)(y-1)=0

y=1, -3

sin(x)=-3 (Can't be, not in [-1,1] interval)

sin(x)=0

x = π/2