Question 1062656
i think you mean cos(theta).
assuming that, i get:


2cos(theta) - sqrt(3) = 0
add sqrt(3) to both sides to get 2cos(theta) = sqrt(3)
divide both sides by 2 to get cos(theta) = sqrt(3)/2.


you want all solutions from 0 to 2pi.


that includes all 4 quadrants from 0 to 360 degrees.


cosine is positive in first and fourth quadrant.


therefore your solution is in the first quadrant and in the fourth quadrant.


it can't be in quadrants 2 and 3 because cosine is negative in those quadrants.


arccosine ((sqrt(3)/2) is equal to 30 degrees.


this is equivalent to 30 * pi / 180 = pi/6 radians.


this is in the first quadrant.


in the fourth quadrant this angle would be equal to 360 - 30 = 330 degrees.


this is equivalent to 330 * pi / 180 = 11pi/6 radians.


i believe your solution is that the angle in radians is equal to 1/6 * pi or 11/6 * pi radians.


that should be equivalent to 30 or 330 degrees


for graphing purposes, you would graph as follows:


graph the equation of y = 2cos(x)


on the same graph, graph the equation of y = sqrt(3).


the intersection of these 2 equations on that graph should be your solution.


alternatively, you should be able to graph y = 2cos(x) - sqrt(3) and your solution should be where y = 0.


the first 2 graphs show the solution in degrees.


the second 2 graphs show the solution in radians.


note that radians = degrees * pi / 180.


not that degrees = radians * 180 / pi.


not that sqrt(3) is equal to 1.732 rounded to 3 decimal digits.


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