Answer:
x = π/3, x = 5π/3, x = 4π/3
Step-by-step explanation:
Let's split the given equation (2cosx-1)(2sinx+√3 ) = 0 into two parts, and solve each separately. These parts would be 2cos(x) - 1 = 0, and 2sin(x) + √3 = 0.
Remember that the general solutions for cos(x) = 1/2 are x = π/3 + 2πn and x = 5π/3 + 2πn. In this case we are given the interval 0 ≤ x ≤2π, and therefore x = π/3, and x = 5π/3.
Similarly:
The general solutions for sin(x) = - √3/2 are x = 4π/3 + 2πn and x = 5π/3 + 2πn. Therefore x = 4π/3 and x = 5π/3 in this case.
So we have x = π/3, x = 5π/3, and x = 4π/3 as our solutions.
Answer:
im in 8th grade and im getting your questions
Step-by-step explanation:
Answer:
(under root 81 is equals to 9
under root 144 is equals to 12
under root 169 is equals to 13) these are rational numbers
therefore irrational number is under root 156
This should be the equation:
n + 4/3n + 5/3n = 90
If one of our zeros is 4, then the factor is x-4. If the second zero is 5i, then the conjugate root theorem says there HAS to be a root that is -5i. So our 3 factors are (x-4)(x+5i)(x-5i). We will FOIL out these factors to get the polynomial. Let's start with the ones that contain the imaginary numbers. Doing that mutliplication we get x^2-25i^2. i^2 is equal to -1, so what that expression simplifies down to is
. Now we will multiply in that last factor of (x-4):
. FOILing out we have
. There you go!
