Solve for 0<x<2pi
---> (2radical2)cos^2(x)-(radical2-2)cos(x)-1=0
To solve the given equation, we can use a quadratic equation formula: ax^2 + bx + c = 0, where a = 2√2, b = -√2 - 2, and c = -1.
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values, we have:
x = [−(-√2 - 2) ± √((-√2 - 2)^2 - 4(2√2)(-1))]/(2(2√2))
Simplifying further:
x = (√2 + 2 ± √(2 - 4(2√2)(-1)))/(4√2)
Next, let's simplify the term inside the square root:
2 - 4(2√2)(-1) = 2 + 8√2 = 2(1 + 4√2)
Now, substitute this back into the equation:
x = (√2 + 2 ± √(2(1 + 4√2)))/(4√2)
The solutions are obtained by setting the expression inside the square root to be greater than or equal to 0:
2(1 + 4√2) ≥ 0
Simplifying:
1 + 4√2 ≥ 0
Since the square root of 2 is a positive number, it is clear that 1 + 4√2 will always be greater than zero. Therefore, we can conclude that there are no solutions in the interval 0 < x < 2π for the given equation.
Hence, the equation (2√2)cos^2(x) - (√2 - 2)cos(x) - 1 = 0 has no solutions in the interval 0 < x < 2π.