2 sin 3θ + 1 = 0?

Question

Find the solutions in the interval [0, 2π).

Answer 1

2 sin3θ + 1 = 0

Given interval is [0 , 2π)

2 sin3θ = -1

sin3θ = -½

sin3θ = sin(-π/6)

  General solution for sin(k) = sin(α) is k = nπ + (-1)ⁿα

  Here k = 3θ , α = -π/6

3θ = nπ + (-1)ⁿ(-π/6)

θ = (π/3)n + (-1)ⁿ(-π/18)

For n = 0 ⇒ θ = (π/3)0 + (-1)⁰(-π/18) = -π/18 is out of given range [0,2π).

For n = 1 ⇒ θ = (π/3)1 + (-1)¹(-π/18)

                     = (π/3)+(-1)(-π/18) = (π/3)+(π/18) = 7π/18

For n = 2 ⇒ θ = (π/3)2 + (-1)²(-π/18)

                     = (π/3)2+(1)(-π/18) = (2π/3)-(π/18) = 11π/18

For n = 3 ⇒ θ = (π/3)3 + (-1)³(-π/18)

                     = (π/3)3+(-1)(-π/18) = (3π/3)+(π/18) = 19π/18

For n = 4 ⇒ θ = (π/3)4 + (-1)⁴(-π/18)

                     = (π/3)4+(1)(-π/18) = (4π/3)-(π/18) = 23π/18

For n = 5 ⇒ θ = (π/3)5 + (-1)⁵(-π/18)

                     = (π/3)5+(-1)(-π/18) = (5π/3)+(π/18) = 31π/18

For n = 6 ⇒ θ = (π/3)6 + (-1)⁶(-π/18)

                     = (π/3)6+(1)(-π/18) = (6π/3)-(π/18) = 35π/18

For n = 7 ⇒ θ = (π/3)7 + (-1)⁷(-π/18)

                     = (π/3)7+(-1)(-π/18) = (7π/3)+(π/18) = 43π/18 is out of given range [0,2π).

So, for n > 7, x will be out of range [0,2π).

Solution :

x = {7π/18 , 11π/18 , 19π/18 , 23π/18 , 31π/18 , 35π/18 }