[0,2pi] 2sin3xx-1=0

Question

solve each trigonometric equation in the interval [0,2pi] 2sin 3x-1=0

Answer 1

2sin(3θ) - 1 = 0

 

2sin(3θ) = 1

 

sin(3θ) = 1/2

 

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

 

θ = π/18 + 2kπ/3, or 5π/18 + 2kπ/3 

If k = 0, then θ = π/18, 5π/18

 

If k = 1, then θ = 13π/18, 17π/18 

If k = 2, then θ = 25π/18, 29π/18

 

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

So there are a total of six solutions, as listed above.