Solve for x, 0<x<2pi
3sin^2x - sinx -2=0
Its not factorable, do I use quadratic formula?
yes, it does factor....
(3sinx + 2)(sinx - 1) = 0
sinx = -2/3 or sinx = 1
if sinx=-2/3, x is in II or III
x = 2.411865 or x = 3.87132
if sinx=1
x = π/2
Quad III and IV?
May I know how you got 3.86132? and value for quad IV?
Yes, if the quadratic equation is not factorable, you can use the quadratic formula to solve for x. The quadratic formula states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In your equation, 3sin^2x - sinx - 2 = 0, the coefficients are:
a = 3, b = -1, and c = -2.
Now, let's substitute these values into the quadratic formula to solve for x:
x = (-(-1) ± √((-1)^2 - 4(3)(-2))) / (2(3))
Simplifying further:
x = (1 ± √(1 + 24)) / 6
x = (1 ± √25) / 6
Now, we consider the two possible solutions:
1. x = (1 + √25) / 6
Simplifying this:
x = (1 + 5) / 6
x = 6/6
x = 1
2. x = (1 - √25) / 6
Simplifying this:
x = (1 - 5) / 6
x = -4/6
x = -2/3
Since you specified that 0 < x < 2π, only the solution x = 1 falls within this range.
Hence, the solution to the equation 3sin^2x - sinx -2 = 0, 0 < x < 2π, is x = 1.