Solve: 2 cos2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi.
To solve the equation 2cos^2x - 3cosx + 1 = 0 for 0 ≤ x < 2π, we can use the quadratic formula or factorization. Let's solve it using factorization method.
First, let's rewrite the equation as follows:
2cos^2x - 3cosx + 1 = 0
Now, we can factorize the quadratic equation. Let's find two numbers a and b such that a*b = 2 and a + b = -3. The numbers -2 and -1 fulfill these conditions.
So, we can rewrite the equation as:
2cos^2x - 2cosx - cosx + 1 = 0
Now we can group the terms:
(2cos^2x - 2cosx) - (cosx - 1) = 0
Factor out the common terms from each group:
2cosx(cosx - 1) - 1(cosx - 1) = 0
Now, we can factor out the common factor (cosx - 1):
(2cosx - 1)(cosx - 1) = 0
Now, we can set each factor equal to zero and solve for x:
1) 2cosx - 1 = 0
2cosx = 1
cosx = 1/2
Using inverse cosine function, we find the solutions for cosx = 1/2:
x = π/3 and x = 5π/3
2) cosx - 1 = 0
cosx = 1
Using inverse cosine function, we find the solution for cosx = 1:
x = 0
So, the solutions to the equation 2cos^2x - 3cosx + 1 = 0, for 0 ≤ x < 2π, are:
x = 0, π/3, and 5π/3.