Find all values of ¸ in [0°, 360°) that satisfy the equation.

2 sin 2θ - √(3) = 0

2sin2θ = √3

sin2θ = √3/2
sin(x) is positive in QI,QII

2θ = 60°,120°,...
we need to go for two periods because we're dealing with 2θ. That gives us

θ = 30°,60°,210°,240°

To find the values of θ that satisfy the equation, we can solve the equation for θ step by step.

The given equation is: 2 sin 2θ - √(3) = 0

Step 1: Divide both sides of the equation by 2 to isolate the sin 2θ term:
sin 2θ - √(3)/2 = 0

Step 2: Move the √(3)/2 term to the right side of the equation:
sin 2θ = √(3)/2

Step 3: Find the reference angle:
The reference angle can be found by taking the inverse sine of √(3)/2, which is a familiar value. The sine of 60° is √(3)/2.

So, the reference angle is 60°.

Step 4: Determine the possible values of 2θ:
Since the sine function has a periodicity of 360°, the equation sin 2θ = √(3)/2 will have two complete cycles within the interval [0°, 360°).

The first cycle occurs when 2θ is equal to the reference angle, 60°:
2θ = 60°
θ = 30°

The second cycle occurs when 2θ is equal to the reference angle plus 180°:
2θ = 60° + 180°
2θ = 240°
θ = 120°

Step 5: Find all values of θ in the interval [0°, 360°):
Since we found two possible values of θ (30° and 120°), we can list all the values of θ that satisfy the equation within the given interval:
θ = 30°, 120°

Therefore, the values of θ that satisfy the equation 2 sin 2θ - √(3) = 0 in the interval [0°, 360°) are 30° and 120°.