Solve:
2cos^2 theta - 3cos theta+ 1= 0
interval [0, 2ð)
let z = cos theta
then
2 z^2 -3 z + 1 = 0
(2z-1)(z-1) = 0
z = 1/2 or z = 1
cos theta = 1/2 gives theta = 60degrees or pi/3 radians and also -60
cos theta = 1 means 0 or 360
To solve the given equation 2cos^2(theta) - 3cos(theta) + 1 = 0 in the interval [0, 2π), we can use factoring or the quadratic formula.
Let's start by factoring the equation:
2cos^2(theta) - 3cos(theta) + 1 = 0
The equation can be factored as follows:
(2cos(theta) - 1)(cos(theta) - 1) = 0
Now, we have two factors:
1) 2cos(theta) - 1 = 0
2) cos(theta) - 1 = 0
To solve these equations separately, let's solve equation 1:
2cos(theta) - 1 = 0
Adding 1 to both sides:
2cos(theta) = 1
Dividing by 2:
cos(theta) = 1/2
To find the solutions for theta, we need to consider the interval [0, 2π). The cosine function has a positive value of 1/2 in the first and fourth quadrants. In these quadrants, the cosine function corresponds to two angles: θ and 2π - θ.
So the solutions for equation 1 are:
θ = arccos(1/2)
θ = 2π - arccos(1/2)
Now let's solve equation 2:
cos(theta) - 1 = 0
Adding 1 to both sides:
cos(theta) = 1
The cosine function has a positive value of 1 at θ = 0 in the interval [0, 2π), so this is another solution.
Therefore, the solutions for the given equation in the interval [0, 2π) are:
θ = arccos(1/2)
θ = 2π - arccos(1/2)
θ = 0