Find all solutions of the equation 3sin^2 x -7sin x +2 = 0 in the interval [0,2\pi).
The answer is x_1= ? and x_2= ? with x_1<x_2.
To find all solutions of the equation 3sin^2 x - 7sin x + 2 = 0 in the interval [0, 2π), we can use the quadratic formula.
The quadratic formula states that if you have an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, the equation 3sin^2 x - 7sin x + 2 = 0 can be rewritten as:
3sin^2 x - 7sin x + 2 = 0
Comparing this equation to the standard quadratic equation form ax^2 + bx + c = 0, we have:
a = 3, b = -7, and c = 2.
Plugging these values into the quadratic formula, we get:
x = (-(-7) ± √((-7)^2 - 4(3)(2))) / (2(3))
Simplifying further:
x = (7 ± √(49 - 24)) / 6
x = (7 ± √25) / 6
x = (7 ± 5) / 6
So the two possible solutions are:
x = (7 + 5) / 6 = 2
x = (7 - 5) / 6 = 2/3
Now we need to check if these solutions fall within the interval [0, 2π). Both solutions do fall within this interval.
Therefore, the solutions for the equation 3sin^2 x - 7sin x + 2 = 0 in the interval [0, 2π) are:
x1 = 2
x2 = 2/3
And since x1 < x2, we have x1 = 2/3 and x2 = 2.