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 = 0y = 4/3
y - 1 = 0y = 1
Now we need to substitute back cosx for y:
cosx = 4/3This solution is not valid because the range of cosx is between -1 and 1.
cosx = 1This 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.
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.