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)
10 sin^2 x - 3sinx - 4 = 0
(2sinx + 1)(5sinx - 4) = 0
sinx = -1/2 or sinx = 4/5
if sinx = -1/2, (x must be in III or IV)
x = 210° or 330°
if sinx = 4/5 , (x must be in I or II)
x = 53.1301 or x = 126.8699°
To solve the equation 10 sin^2 x = 3 sin x + 4 in the interval [0, 2π), we can use the inverse trigonometric functions. The equation can be rewritten as a quadratic equation by moving all the terms to one side:
10 sin^2 x - 3 sin x - 4 = 0
Now, let's introduce a new variable, t, to simplify the equation:
t = sin x
Replacing sin x with t, we get:
10t^2 - 3t - 4 = 0
Now we can solve this quadratic equation for t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In our case, a = 10, b = -3, and c = -4. Plugging these values into the quadratic formula:
t = [-(-3) ± √((-3)^2 - 4 * 10 * (-4))]/(2 * 10)
= (3 ± √(9 + 160))/20
= (3 ± √(169))/20
Simplifying and evaluating the square root:
t = (3 ± 13)/20
= 16/20 or -10/20
= 4/5 or -1/2
Now we need to find the corresponding values of x for each value of t. Since t = sin x, we can use the inverse sine function (sin^-1) to solve for x:
x = sin^-1 (t)
For t = 4/5:
x = sin^-1 (4/5) ≈ 0.9273
For t = -1/2:
x = sin^-1 (-1/2) ≈ -0.5236 or 2π - 0.5236 ≈ 5.7596
In the interval [0, 2π), the solutions to the equation are approximately:
x ≈ 0.9273, -0.5236, 5.7596
Therefore, the solutions in the given interval [0, 2π) are 0.9273, -0.5236, and 5.7596.
To find the solutions of the equation in the given interval using inverse trigonometric functions, we need to rewrite the equation in terms of the inverse functions. In this case, we'll use the inverse of the sine function, which is arcsin or sin^(-1).
The given equation is:
10 sin^2(x) = 3 sin(x) + 4
First, we rewrite the equation by moving all the terms to one side:
10 sin^2(x) - 3 sin(x) - 4 = 0
Next, we replace sin(x) with a variable, let's say u:
10 u^2 - 3 u - 4 = 0
Now, we can solve this quadratic equation for u using factoring or the quadratic formula. Factoring doesn't seem to work easily in this case, so we'll use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 10, b = -3, and c = -4. Substituting these values into the quadratic formula, we have:
u = (-(-3) ± √((-3)^2 - 4(10)(-4))) / (2(10))
u = (3 ± √(9 + 160)) / 20
u = (3 ± √169) / 20
u = (3 ± 13) / 20
Now, we have two possible values for u:
u₁ = (3 + 13) / 20 = 16 / 20 = 0.8
u₂ = (3 - 13) / 20 = -10 / 20 = -0.5
Since we replaced sin(x) with u, we can find x using the inverse sine (arcsin) function:
x = arcsin(u)
To find the solutions within the interval [0, 2π), we need to check both values of u and determine their corresponding values of x.
For u = 0.8:
x₁ = arcsin(0.8) ≈ 0.9273 (in radians)
For u = -0.5:
x₂ = arcsin(-0.5) ≈ -0.5236 (in radians)
Note: We use radians as the unit of measurement to find the angles within the specified interval.
Finally, we can approximate the solutions to four decimal places and write them as a comma-separated list:
x₁ ≈ 0.9273, x₂ ≈ -0.5236