What are the values of theta in the interval 0 less than or equal to theta is less than 360 that sastify the equation sin theta - 1/2 =0?
sinØ - 1/2 = 0
sinØ = 1/2
so Ø is in quadrants I or II
we know sin 30° = 1/2
so Ø = 30° or 150° , ( π/6 or 5π/6 radians)
To find the values of theta that satisfy the equation sin(theta) - 1/2 = 0 in the interval 0 ≤ θ < 360, we need to solve the equation for theta.
First, let's isolate the sine function:
sin(theta) - 1/2 = 0
Add 1/2 to both sides:
sin(theta) = 1/2
Now, we know that the sine function is positive in two quadrants: the first quadrant (0° ≤ θ ≤ 90°) and the second quadrant (90° < θ ≤ 180°).
We also know that sin(theta) = 1/2 at two corresponding angles: 30° and 150°. These are known as the "reference angles."
To find the values of theta that satisfy sin(theta) = 1/2, we can use the reference angles as a starting point and consider the symmetrical nature of the sine function.
Start with the first quadrant reference angle, 30°. Since sine is positive in both the first and second quadrants, we add the angle of rotation in the second quadrant (180°) to the reference angle:
θ = 30° + 180° = 210°
So, 210° is one value of theta that satisfies sin(theta) - 1/2 = 0.
Next, we move to the second quadrant reference angle, 150°. Again, we add the angle of rotation in the second quadrant (180°) to the reference angle:
θ = 150° + 180° = 330°
Therefore, 330° is the other value of theta that satisfies sin(theta) - 1/2 = 0.
In summary, the values of theta in the interval 0 ≤ θ < 360 that satisfy the equation sin(theta) - 1/2 = 0 are 210° and 330°.