SOLUTION: Find all solutions
6 cos 2θ − 3 square root 3= 0
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Question 863138: Find all solutions
6 cos 2θ − 3 square root 3= 0
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
equation is:
6cos(2x) - 3sqrt(3) = 0
add 3sqrt(3) to both sides of the equation to get:
6cos(2x) = 3sqrt(3)
divide both sides by 6 to get:
cos(2x) = 3sqrt(3)/6 = sqrt(3)/2
you get 2x = 30 as a solution.
that would be in quadrant 1.
cosine is positive in quadrant 1 and quadrant 4.
your solutions are:
2x1 = 30
2x2 = 330
that cycle repeats every 360 degrees for each one of these solutions.
your solution are:
2x1 = 30 +/- 360*k
2x2 = 330 +/- 360*k
k is a non-negative integer.
it can go from 0 to infinity.
if you want to solve for x, then you need to divide both sides of this equation by 2 to get:
2x1 = (30 +/- 360*k) / 2
2x2 = (330 +/- 360*k) / 2
you can graph the equation of 6cos(2x) = 3sqrt(3) = y and your solutions will be at the zero crossing points.
since the graph assumes radians, then the solutions will be at:
x1 = (pi/6 +/- 2pi*k)/2
x2 = (11pi/6 +/- 2pi*k)/2
translation from degrees to radians is:
radians = degrees * pi / 180
30 * pi / 180 = pi / 6
330 * pi / 180 = 11pi / 6
graph looks like this:
some check points are:
x1 = (pi/6)/2 = approximately .27
x1 = (pi/6 + 2pi*1)/2 = approximately 3.4
x1 = (pi/6 + 2pi*2)/2 = approximately 6.5
x2 = (11pi/6)/2 = approximately 2.9
x2 = (11pi/6 + 2pi*1)/2 = approximately 6.0
x2 = (11pi/6 + 2pi*2)/2 = approximately 9.2
you can see from the graph that these are approximate zero crossing points.
you can calculate other values from the formulas.
those formulas are:
x1 = (pi/6 +/- 2pi*k)/2
x2 = (11pi/6 +/- 2pi*k)/2
those formulas are your solution to this problem.
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