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, we can use inverse trigonometric functions.
Let's rewrite the equation in terms of sine function:
10 sin^2 x = 3 sin x + 4
To apply inverse trigonometric functions, we need to isolate sin x. Rearranging the equation, we have:
10 sin^2 x - 3 sin x - 4 = 0
Let's define y = sin x. The equation becomes:
10 y^2 - 3 y - 4 = 0
Now, we can solve this quadratic equation for y. We can either factor it or use the quadratic formula. However, in this case, factoring seems more challenging compared to using the quadratic formula.
The quadratic formula is given by:
y = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our equation, a = 10, b = -3, and c = -4. Substituting these values into the quadratic formula, we get:
y = (3 ± sqrt((-3)^2 - 4(10)(-4))) / (2(10))
y = (3 ± sqrt(9 + 160)) / 20
y = (3 ± sqrt(169)) / 20
y = (3 ± 13) / 20
Now, we have two possible values for y:
1. y = (3 + 13) / 20 = 16 / 20 = 0.8
2. y = (3 - 13) / 20 = -10 / 20 = -0.5
Now, using the inverse trigonometric functions, we can find the values of x in the interval [0, 2π) corresponding to these values of y.
The first solution is:
sin x = 0.8
Using the inverse sine function (sin^(-1)) or arcsin, we find:
x = sin^(-1)(0.8)
Using a calculator or table of values for the inverse sine function, we can approximate this value to four decimal places:
x ≈ 0.9273
The second solution is:
sin x = -0.5
Using the inverse sine function (sin^(-1)) or arcsin, we find:
x = sin^(-1)(-0.5)
Again, using a calculator or table of values for the inverse sine function, we can approximate this value to four decimal places:
x ≈ 4.7124
Therefore, the solutions in the interval [0, 2π) are approximately x ≈ 0.9273 and x ≈ 4.7124.