Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. (Enter your answers as a comma-separated list.)

10 sin^2 x = 3 sin x + 4; [0, 2π)

To find the solutions of the equation 10sin^2 x = 3sin x + 4 in the interval [0, 2π), we can use inverse trigonometric functions.

1. Rewrite the equation: 10sin^2 x - 3sin x - 4 = 0.

2. Let's introduce a substitution to simplify the equation. Let t = sin x.

The equation becomes: 10t^2 - 3t - 4 = 0.

3. Solve this quadratic equation using the quadratic formula:

t = (-(-3) ± √((-3)^2 - 4 * 10 * -4)) / (2 * 10)
= (3 ± √(9 + 160)) / 20
= (3 ± √169) / 20
= (3 ± 13) / 20

We have two possible solutions for t: t1 = (3 + 13) / 20 and t2 = (3 - 13) / 20.

4. Calculate the values of t:

t1 = 16 / 20 = 4/5
t2 = -10 / 20 = -1/2

5. Now, we can find the values of x using the inverse sine function (sin^-1):

x1 = sin^-1(4/5)
x2 = sin^-1(-1/2)

6. To find the solutions within the interval [0, 2π), we need to consider the periodicity of the sine function.

Since sin x has a period of 2π, we can add integer multiples of 2π to the solutions.

For x1:
- x1 is in the first quadrant, so it will have the same value within the interval [0, 2π).
x1 ≈ 0.9273

For x2:
- x2 is in the fourth quadrant, so we need to calculate its reference angle.
- The reference angle is sin^-1(1/2), which is π/6 (approximately 0.5236).
- To get the angle in the fourth quadrant, we subtract the reference angle from 2π.
x2 ≈ 2π - π/6 ≈ 11π/6 ≈ 5.7596

7. Combine the solutions within the interval [0, 2π):

The solutions are approximately x1 ≈ 0.9273, x2 ≈ 5.7596.

Therefore, the solutions to the equation within the interval [0, 2π) are approximately 0.9273 and 5.7596.