Solve the equation...

Question

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

Answer 1

In order to solve an equation like this one, the first thing we want to do is get everything in terms of the same angle, since right now we have a 2θ on one side and a θ on the other and those cannot be compared directly. The double angle identity lets us fix that, because sin(2θ) is the same thing as 2sin(θ)cos(θ), so the equation becomes:

2sin(θ)cos(θ) = cos(θ)

Now here is the step where a lot of students quietly lose two of the four solutions. It is tempting to divide both sides by cos(θ), but doing that assumes cos(θ) is not zero, and the cases where it actually is zero are real solutions you would be throwing away. So instead of dividing, we move everything to one side and factor:

cos(θ)(2sin(θ) - 1) = 0

A product is zero when either piece is zero, so now we just handle the two pieces separately. From cos(θ) = 0 we get θ = π/2 and θ = 3π/2, and from 2sin(θ) - 1 = 0 we get sin(θ) = 1/2, which happens at θ = π/6 and θ = 5π/6 on the interval we were given.

So, our four solutions are θ = π/6, π/2, 5π/6, and 3π/2, which lines up with the hint that there would be four of them.

Answer 2

 

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 3

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