We are given the equation 1+sin8 = 1/2 and we need to find its solution on the interval [0,2π).
To solve this equation, we will start by isolating the sine term.
1 + sin8 = 1/2
Subtracting 1 from both sides, we get:
sin8 = -1/2
Now, we need to find the angles on the interval [0,2π) that have a sine of -1/2.
The sine function is negative in the third and fourth quadrants of the unit circle. Therefore, we will look for angles in these quadrants that have a sine of 1/2.
To find these angles, we will use the fact that the sine function is equal to the y-coordinate of the point on the unit circle.
Let's start with the third quadrant. In this quadrant, the x-coordinate is negative and the y-coordinate is negative. Therefore, we can write:
sinθ = -1/2 = sin(π - θ)
This gives us:
π - θ = arcsin(-1/2) = -π/6
θ = π + π/6 = 7π/6
Now, let's move on to the fourth quadrant. In this quadrant, both the x-coordinate and the y-coordinate are positive. Therefore, we can write:
sinθ = -1/2 = sin(2π - θ)
This gives us:
2π - θ = arcsin(-1/2) = -π/6
θ = 2π + π/6 = 11π/6
Therefore, the solutions of the given equation on the interval [0,2π) are:
A. 7π/6, 11π/6
And the correct option is A.