Final answer:
To solve 2 sin(3θ) − 1 = 0, find the inverse sine of ½, use the sine function's periodicity to find all angles, and divide by 3 to find θ. The solutions are given by θ = 0.167 + 2.094k and θ = 0.977 + 2.094k in radians, where k is any integer.
Step-by-step explanation:
To find all solutions to the equation 2 sin(3θ) − 1 = 0, follow these steps:
- Isolate the trigonometric function: sin(3θ) = ½.
- Find the principal value of the angle 3θ using the inverse sine function: 3θ = sin⁻¹(½).
- Use the symmetry and periodicity of the sine function to find all possible angles.
- Since the sine function has a period of 2π, add multiples of 2π to the principal value: 3θ = sin⁻¹(½) + 2πk, where k is any integer.
- Also, because sine is positive in the first and second quadrants, there is a second set of solutions: 3θ = π - sin⁻¹(½) + 2πk.
- Divide by 3 to find all solutions for θ: θ = sin⁻¹(½)/3 + 2πk/3 and θ = (π - sin⁻¹(½))/3 + 2πk/3.
- Calculate the numerical values of the solutions to three decimal places, in radians.
The solutions of the equation are therefore θ = 0.167 + 2.094k and θ = 0.977 + 2.094k, where k is any integer.
It's important to check whether these solutions make sense: the initial equation is a basic trigonometric equation, and the periodic nature of the sine function means we should expect an infinite set of solutions spaced out by the function's period, which is correctly reflected in the answers.