SOLUTION: Find all solutions in the interval [0,2pi) 4cos(theta)=1+2cos(theta)

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Question 1206253: Find all solutions in the interval [0,2pi)
4cos(theta)=1+2cos(theta)
Found 2 solutions by MathLover1, Theo:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

4cos%28theta%29=1%2B2cos%28theta%29
4cos%28theta%29-2cos%28theta%29+=1
2cos%28theta%29+=1
cos%28theta%29+=1%2F2
theta+=cos%5E-1%281%2F2%29
theta+=pi%2F3

solutions in the interval [0,2pi)
theta+=pi%2F3
theta=+5pi%2F3


Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let x = theta.
equation is 4cos(x) = 1 + 2cos(x)
subtract 2cos(x) from both sides of the equation to get:
2cos(x) = 1
divide both sides of the equation by 2 to get:
cos(x) = 1/2
cosine is positive in quadrants 1 and 4.
in quadrant 1, x = arccos(1/2) = pi/3.
the equivalent angle in the fourth quadrant is 2pi - pi/3 = 5pi/3.
the graph of both equations is shown in below.
4cos(x) = 1 + 2cos(x) where the graph of both equations intersect in the interval (0,2pi).



translating to degrees.
0 radians * 180/pi = 0 degrees.
2pi radians * 180/pi = 360 degrees.
pi/3 radians * 180/pi = 60 degrees.
5pi/3 radians * 180/pi = 300 degrees.