Question

Solve the trigonometric equation for all values 0 < x < 5 for sin(3x) = 0. What are the solutions?

a) x = 0, x = π, x = 2π
b) x = π/3, x = 2π/3, x = π, x = 2π/3
c) x = π/3, x = π, x = 2π/3
d) x = π/3, x = 2π/3, x = π, x = 4π/3

Answer 1

The trigonometric equation sin(3x) = 0 for 0 < x < 5 is solved by considering when the sine function equals zero, which is at multiples of π. The solutions within the range are x = π/3, 2π/3, π, and 4π/3.

Step-by-step explanation:

To solve the trigonometric equation sin(3x) = 0 for all values 0 < x < 5, we need to consider the basic property of the sine function that it is zero at points of the form nπ, where n is an integer. Since we have sin(3x), we need the argument 3x to be equal to nπ, hence 3x = nπ. Solving for x gives us x = nπ/3.

Now, we look for all multiples of π/3 that fall within the given range (0, 5). The possible values for n that provide solutions in the given interval are 1, 2, 3, and 4, which correspond to x = π/3, 2π/3, π, and 4π/3, respectively. These values are within the specified range and are therefore solutions to the equation.

Thus, considering the given range, the complete solution to the trigonometric equation sin(3x) = 0 for 0 < x < 5 is x = π/3, x = 2π/3, x = π, and x = 4π/3.