solve the equation on the interval 0 = ? 2p

Question

2sin²θ=sinθ

Answer 1

• 2sin²θ - sinθ = 0
• sinθ (2sinθ - 1) = 0 ~ Factor
• Now you can use the 0 principle, where each piece can equal 0, so you are looking at two sets of possible solutions

 

1. sinθ = 0

• θ = 0 or π (pi...that symbol doesn't look like it says it will)

2. 2sin θ -1 = 0

• 2sin θ = 1
• sin θ = 1/2
• θ = π/6 and 5π/6

 

So you have four solutions in the range 0<θ<2π, and those solutions are 0, π/6, 5π/6, and π .  (Note if your range does not include the endpoints of 0 and/or 2π, exclude those accordingly.)