Solve the equation on the interval 0≤θ<2π. cos(2θ)=− sqrt3/2

Answer 1

cos(2θ) = -√3/2

So, 2θ = 5π/6 + 2kπ or 7π/6 + 2kπ, where k = 0, 1, -1, 2, -2, ... (recall that cosx repeats every 2π).

Therefore, θ = 5π/12 + kπ or 7π/12 + kπ

If k = 0, then θ = 5π/12 or 7π/12

If k = 1, then θ = 17π/12 or 19π/12

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