Solve sin^2x - 2sinx - 3 = 0 on the interval xE [0, 2pie]
pi, not pie
sin^2x - 2sinx - 3 = 0
(sinx-3)(sinx+1) = 0
sinx=3 nope
sinx = -1
x = 3pi/2
4cos2x - cosx-3 =0
0 to 2pi
To solve the equation sin^2x - 2sinx - 3 = 0 on the interval xE [0, 2π], we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, we have a = 1, b = -2, and c = -3. Substituting these values into the quadratic formula, we get:
x = (-(-2) ± √((-2)^2 - 4(1)(-3))) / (2(1))
= (2 ± √(4 + 12)) / 2
= (2 ± √16) / 2
= (2 ± 4) / 2
So, we have two possible solutions:
1. x = (2 + 4) / 2 = 6 / 2 = 3
2. x = (2 - 4) / 2 = -2 / 2 = -1
Now, we need to check if these solutions lie within the given interval [0, 2π].
For x = 3:
Since 3 is within the interval [0, 2π], we keep this as one of the solutions.
For x = -1:
Since -1 is outside the interval [0, 2π], we discard this solution.
Therefore, the only solution within the given interval is x = 3.
To solve the equation sin^2x - 2sinx - 3 = 0 on the interval x ∈ [0, 2π], we can use a quadratic equation approach.
First, let's make a substitution to simplify the equation: let u = sin(x). Now, the equation becomes u^2 - 2u - 3 = 0.
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula in this case.
The quadratic formula states that for 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).
Applying this formula to our equation, we have:
u = (-(-2) ± √((-2)^2 - 4(1)(-3))) / (2(1))
u = (2 ± √(4 + 12)) / 2
u = (2 ± √16) / 2
u = (2 ± 4) / 2
This gives us two possible solutions for u:
1) (2 + 4) / 2 = 6 / 2 = 3
2) (2 - 4) / 2 = -2 / 2 = -1
Now, we need to find the corresponding values of x for these solutions within the specified interval [0, 2π].
For u = 3, we can find the corresponding x-value by solving sin(x) = 3. However, this is impossible since the range of the sine function is -1 to 1. Therefore, there are no solutions for this case.
For u = -1, we can find the corresponding x-value by solving sin(x) = -1. This occurs when x = 3π/2.
Therefore, the only solution within the interval [0, 2π] is x = 3π/2.