R = q - r
2r = q
r = q\2
answer to the question
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
1. 102°
2. 56°
3. 104°
4. 138°
5. 95°
6. 133°
Answer:
Option 4
Step-by-step explanation:
-½ means it's -0.5, and then -0.4 then -3/10 = -0.3 and then -0.23, therefore it goes -½, -0.4, -3/10, then -0.23