Solve each trigonometric equation in the interval [0,2π). 2 sin3x-1=0

Question

please help me understand this!!

Answer 1

Hi Tetyana,

 

2 sin3x-1=0

2sin3x = 1

sin3x = 1/2

3x = 30, 150, 390, 510, 750, 870 degrees (these will satisfy for x across [0, 2pi))

x ={10, 50, 130, 170, 250, 290)

 

 

 

Answer 2

sin 3x - 1 = 0

 

sin 3x = 1

 

3x = 2kπ + π/2 where k ∈ Z

 

x = 2kπ/3 + π/6 where k ∈ Z

 

In the [0,2π), the solutions are π/6, 5π/6, and 3π/2.