How would I go about solving this equation on the interval [0,2pi]
cos^2 + 2 cos x + 1 = 0
cos^2(x) + 2 cos(x) + 1 = [cos(x) + 1]^2
So, cos(x) = -1 ------>
x = (2n+1)pi
Muliple choice questions I have 4 options
answers.
1. pi/4, 7pi/4
2. pi/2, 3pi/2
3. pi
4. 2pi
What formula do I use ?
thanks
To solve the equation cos^2(x) + 2 cos(x) + 1 = 0 on the interval [0, 2pi], we can follow these steps:
Step 1: Rearrange the equation in a quadratic form.
Replace cos^2(x) with (cos(x))^2 to get (cos(x))^2 + 2cos(x) + 1 = 0.
Step 2: Simplify the equation.
Combine like terms to get (cos(x))^2 + 2cos(x) + 1 = 0.
Step 3: Factor the equation.
Since the equation is in quadratic form, we can try factoring it. However, this quadratic equation cannot be factored easily. Instead, we can use a substitution technique.
Step 4: Substitute a new variable.
Let u = cos(x). Now, the equation becomes u^2 + 2u + 1 = 0.
Step 5: Solve the quadratic equation.
Factor the quadratic equation (u + 1)^2 = 0, which gives u = -1.
Step 6: Back-substitute.
Since u = cos(x), we substitute -1 back into the equation to find cos(x) = -1.
Step 7: Find the solutions within the given interval.
To find the solutions within the interval [0, 2pi], we need to check where cos(x) = -1 lies on the unit circle. The solutions should be at x = pi.
Therefore, the solution to the equation cos^2(x) + 2cos(x) + 1 = 0 on the interval [0, 2pi] is x = pi.