Find all real numbers in the interval [0,2pi) that satisfy the equation.
cos^2 x+2 cos x+1 = 0
let z = cos x
then
z^2 + 2 z + 1 = 0
(z+1)(z+1) = 0
z = -1
so
cos x = -1
that is at x = 180
You can convert those degrees to radians I trust.
To find the real numbers in the interval [0, 2π) that satisfy the equation cos^2(x) + 2cos(x) + 1 = 0, we can solve the quadratic equation.
Let's rewrite the equation as follows:
(cos(x))^2 + 2cos(x) + 1 = 0
This equation can be factored as:
(cos(x) + 1)(cos(x) + 1) = 0
From this, we see that the only way for the product of two terms to be equal to zero is if at least one of them is equal to zero. Therefore, we have two cases to consider:
1. cos(x) + 1 = 0
2. cos(x) + 1 = 0
To find the solutions for each case, let's solve the equations:
1. cos(x) + 1 = 0
Subtracting 1 from both sides:
cos(x) = -1
The value of cos(x) equals -1 when x is π radians or 180 degrees.
2. cos(x) + 1 = 0
Subtracting 1 from both sides:
cos(x) = -1
The value of cos(x) equals -1 when x is 3π radians or 540 degrees.
Thus, the real numbers in the interval [0, 2π) that satisfy the equation cos^2(x) + 2cos(x) + 1 = 0 are π and 3π.