Solve sin(2θ)=−√3/2 for θ in the interval [ 0 , 2 π )

Question

For the first four solutions

Answer 1

sin(2θ) = -√3/2

2θ = 4π/3 + 2kπ or 5π/3 + 2kπ

So, θ = 2π/3 + kπ or 5π/6 + kπ, k = 0, ±1, ±2, ...

If k = 0, we get θ = 2π/3, 5π/6

If k = 1, we get θ = 5π/3, 11π/6

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