Question

Find all values of θ in the interval ​[0°​,360°​) that have the

given function value.
Tan θ = square root of 3 over 3

Answer 1

Values of θ in the interval [0°, 360°) where tan θ equals the square root of 3 over 3 are 30° and 210°.

Step-by-step explanation:

To find all values of θ in the interval [0°, 360°) where tan θ = √3 / 3, we first recognize that √3 / 3 is √3 divided by 3, which simplifies to tan 30° or tan π/6. This value is positive, so θ could be in the first or third quadrant. To find these values, we use the fact that the tangent function has a period of 180° (or π radians). So, we take the principal value θ = 30° and add 180° to find the other value within the specified range.

Therefore, the solutions are θ = 30° (first quadrant) and θ = 30° + 180° = 210° (third quadrant).

Answer 2

To find all values of θ in the interval [0°, 360°) that have the given function value of tan θ = √3/3, we need to determine the angles in which the tangent of that angle equals √3/3. The solutions are θ = 30°, 150°, 390°, 510°, 750°, 870°, ...

Step-by-step explanation:

To find all values of θ in the interval [0°, 360°) that have the given function value of tan θ = √3/3, we need to determine the angles in which the tangent of that angle equals √3/3. Since the tangent of an angle is equal to the ratio of the opposite side to the adjacent side of a right triangle, we can use the special triangles (30-60-90 and 45-45-90) to find the values of θ.

The tangent of 30° is equal to √3/3, so one solution is θ = 30°.

The tangent of 150° is also equal to √3/3, so another solution is θ = 150°.

Since the interval is [0°, 360°), we can also add or subtract multiples of 360° to these solutions. Therefore, the complete set of solutions is:

θ = 30°, 150°, 390°, 510°, 750°, 870°, ...