Please help with this trig question!

Question

Solve: cos(2θ)=-1/2 on the interval 0≤θ<2π

Answer 1

If cos(2θ) = -1/2, then 2θ = 2π/3 + 2kπ or 4π/3 + 2kπ, k = 0, ±1, ±2, ...

 

So, θ = π/3 + kπ, or 2π/3 + kπ

 

If k = 0, we get θ = π/3, 2π/3

 

If k = 1, we have θ = 4π/3, 5π/3

 

Other choices of k give values of θ that lie outside of the interval [0, 2π).

 

There are 4 solutions:  π/3, 2π/3, 4π/3, 5π/3  

 

Answer 2

cos is negative in the 2nd and 3rd quadrants

look on the unit circle

cos(2∏/3)=-1/2     (2∏/3=120º)

so 2θ=120º

      θ=60º

360º-120º=240º and 240º=4∏/3

cos(4∏/3)=-1/2

so 2θ=240º

      θ=120º

θ=60º and 120º

Answer 3

cosine of (2θ) equals negative one half for 2θ = (2/3)pi + 2k(pi)  or   (4/3)pi + 2k(pi).

 

Solving for θ we have pi/3, 2pi/3, 4pi/3