find all solutions of the equation 4cos^2x-4cosx+1=0
Let cos x = y and solve the resulting equation
4y^2 -4y +1 = 0
That equation can be rewritten
(2y-1)^2 = 0
The only solution is
y = cos x = 1/2
x = cos^-1 1/2
Which angles x have cos x = 1/2 ?
There are two solutions between x = 0 and x = 2 pi
To find all solutions of the equation 4cos^2x - 4cosx + 1 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In the given equation, we have a = 4, b = -4, and c = 1. Substituting these values into the quadratic formula, we get:
x = (-(-4) ± √((-4)^2 - 4 * 4 * 1)) / (2 * 4)
x = (4 ± √(16 - 16)) / 8
x = (4 ± √0) / 8
Since the square root of 0 is 0, we have:
x = (4 ± 0) / 8
This simplifies to:
x = 4 / 8
x = 1/2
Therefore, the equation 4cos^2x - 4cosx + 1 = 0 has a single solution, which is x = 1/2.
To find all the solutions of the equation 4cos^2x - 4cosx + 1 = 0, we'll solve it step-by-step.
Step 1: Let's denote cos(x) as "t". Therefore, the equation becomes:
4t^2 - 4t + 1 = 0.
Step 2: Now, we can try to factorize the quadratic equation:
(2t - 1)(2t - 1) = 0.
Step 3: Simplifying the factored form, we get:
(2t - 1)^2 = 0.
Step 4: Taking the square root of both sides:
2t - 1 = 0.
Step 5: Solving for t, we have:
2t = 1,
t = 1/2.
Step 6: Since we let t = cos(x), we can substitute back to find x:
cos(x) = 1/2.
Step 7: To find all the solutions for x, we need to consider the unit circle. From the unit circle, the values for which cos(x) = 1/2 are x = π/3 + 2πn and x = 5π/3 + 2πn, where n is an integer.
Therefore, the solutions of the equation 4cos^2x - 4cosx + 1 = 0 are:
x = π/3 + 2πn and x = 5π/3 + 2πn, where n is an integer.