Solve the equation...

Question

Solve sin(2θ) = cos(θ) for 0 ≤ θ < 2π (Hint: there will be a total of 4 solutions!)

Answer 1

 

Hi Erin,

 

    sin(2θ) = cos(θ), 0 ≤ θ ≤ 2π

 

Acoording to the identy sin(2θ) = 2sin(θ)cos(θ),

 

    2sin(θ)cos(θ) = cos(θ)

 

    2sin(θ)cos(θ) - cos(θ) = 0

 

    cos(θ)(2sin(θ) - 1) = 0

 

    So, cos(θ) = 0 or sin(θ) = 1/2.

 

    cos(θ) = 0 → θ = π/2, 3π/2.

 

    sin(θ) = 1/2 → θ = π/6, 5π/6.

 

The answer includes π/6, π/2, 5π/6, and 3π/2.

 

 

 

Answer 2

sin(2θ) = cos(θ)

 2sinθcosθ-cosθ=0

cosθ(2sinθ-1)=0

cosθ=0 or sinθ=1/2

so θ= pi/2, 3pi/2, pi/6,5pi/6