In place of t, or theta, I'm going to utilize x instead. So the equation is -3*cos(x) = 1. Get everything to one side and we have -3*cos(x)-1 = 0
Let f(x) = -3*cos(x)-1. The goal is to find the root of f(x) in the interval [0, 2pi]
I'm using the program GeoGebra to get the task done of finding the roots. In this case, there are 2 roots and they are marked by the points A and B in the attachment shown
A = (1.91, 0)
B = (4.37, 0)
So the two solutions for theta are
theta = 1.91 radians
theta = 4.37 radians
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Answer:
- 1/64 (x+4)² (x-3) (x-4) = 0
Step-by-step explanation:
f(x) = a (x+4)²(x-3)(x-4) ... curve touch (-4,0) and not cross x axis
x=0 f(x) = -3
a (4)² * (-3) * (-4) = -3
a = - 1/64
- 1/64 (x+4)² (x-3) (x-4) = 0
I think the answer is 2.The function is negative for all real values of x where x < –3 and where x > 1.