Find all solutions of the equation 2sin^2 x + 3sinx+1 = 0
over the interval [0,2pi)
To find the solutions of the given equation over the interval [0, 2π), we can use the quadratic formula since the equation is quadratic in terms of sin(x).
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using:
x = (-b ± √(b^2 - 4ac))/(2a)
In our case, let's rewrite the equation as follows:
2sin^2 x + 3sin x + 1 = 0
This equation is already in the form ax^2 + bx + c = 0 with a = 2, b = 3, and c = 1. Now, we can apply the quadratic formula to solve for sin(x):
sin(x) = (-b ± √(b^2 - 4ac))/(2a)
sin(x) = (-3 ± √(3^2 - 4(2)(1)))/(2(2))
sin(x) = (-3 ± √(9 - 8))/4
sin(x) = (-3 ± √1)/4
sin(x) = (-3 ± 1)/4
Now we have two possibilities:
1. sin(x) = (-3 + 1)/4 = -2/4 = -1/2
2. sin(x) = (-3 - 1)/4 = -4/4 = -1
To find the values of x for which sin(x) equals each of these two possibilities, we can use the inverse sine function (sin^(-1)).
For sin(x) = -1/2:
x = sin^(-1)(-1/2)
Using a calculator, we find that sin^(-1)(-1/2) equals -π/6 (or -30 degrees) and 7π/6 (or 210 degrees).
For sin(x) = -1:
x = sin^(-1)(-1)
Again using a calculator, we find that sin^(-1)(-1) equals -π/2 (or -90 degrees).
Therefore, the solutions to the equation 2sin^2 x + 3sin x + 1 = 0 over the interval [0, 2π) are:
x = -π/6, -π/2, 7π/6
(2sinx +1)(sinx +1) = 0
2sinx + 1 = 0 sinx + 1 = 0
sinx = -1/2 sinx = -1
Can you find x? There are two possible values for the first part and one possible value for the second part given the interval.