To solve this quadratic equation, let's first substitute y for cosx:

3y^2 - 7y + 4 = 0

Now, we need to find the factors of the quadratic equation:

(3y - 4)(y - 1) = 0

Setting each factor to zero and solving for y, we get:

3y - 4 = 0
y = 4/3

y - 1 = 0
y = 1

Now we need to substitute back cosx for y:

cosx = 4/3
This solution is not valid because the range of cosx is between -1 and 1.

cosx = 1
This solution is valid, because the cosine of an angle cannot exceed 1.

Therefore, the solution to the equation is cosx = 1. This means that x equals 0 or any multiple of 2π.