Question

Solve the equation: 2 sin² θ = sin θ, θ ∈(0, π)

Answer 1

Answer: The answer is π/3 & 2π/3

Given that

2 sin² θ = sin θ,                   where θ ∈(0, π)

Now we take

2 sin² θ = sin θ

2 sin² θ - sin θ = 0

sin θ ( 2 sin θ - 1 ) = 0

⇒  sin θ = 0      AND     2 sin θ - 1  = 0

For sin θ = 0 :

   sin θ = 0  

⇒   θ = 0   and θ = π

But given that θ ∈(0, π) ; Which is an OPEN Interval.

0 ∉ (0, π)   And  π ∉ (0, π)

So sin θ = 0  has NO solution.

For 2 sin θ - 1 = 0 :

2 sin θ - 1 = 0

2 sin θ = 1

sin θ = 1 / 2

The Reference Angle is π/3

Sin is + ve in Quadrant I & II.

For Quad I:-  θ = π/3

For Quad II:- θ = π - π/3 = 2π/3

The answer is π/3 & 2π/3