f ′(t)  =  −18 sin(t) + 18 cos(2t) =  −18(2 sin2(t) + sin(t) − 1),which factors to f ′(t) =

The given equation is:

f′(t) = −18 sin(t) + 18 cos(2t) = −18(2 sin^2(t) + sin(t) - 1)

We need to factorize the equation on the right side of the equal sign.

First, we can factor out -18 from each term inside the parentheses:

f′(t) = -18[2sin^2(t) + sin(t) - 1]

Next, we can rearrange the terms inside the parentheses to form a quadratic equation in terms of sin(t):

f′(t) = -18[2sin^2(t) + sin(t) - 1] = -18[(2sin(t))^2 + 2sin(t) - 1]

Now, we can factorize the quadratic equation inside the parentheses. This can be done by finding two numbers that multiply to -1 (the constant term) and add to 2 (the coefficient of the linear term).

The numbers that satisfy these conditions are -1 and 1. Therefore, the factorization of the quadratic equation is (2sin(t) - 1)(sin(t) + 1).

So, the factorized form of the given equation is:

f′(t) = -18(2sin(t) - 1)(sin(t) + 1)