Find all solutions of the equation in the interval[0, 2π)

Question

4cos3x−2√3=0

Answer 1

agree that cos(3x)=(√3)/2

but 3x now should be 3x=cos^-1((√3)/2)+2πk, where yes cos^-1((√3)/2)=+/- π /6 radiands (+/- 30 degrees)

so x=+/-π /18 + (2π k)/3, K=0,1,2 since k=4 same as k=0,k=5 same as k=1, etc.

between [0.2π ) there are these six values for x, ((π /18).(11*π /18),(13*π /18),(23* π /18), and (25*π /18))

remember that it is 3x in the equation so all the denominations become 6.

you can plug in these values for x and check each one has cos3x=.866 =(√3)/2

Answer 2

4cos(3x) - 2√3 = 0

4cos(3x) = 2√3

cos(3x) = √3/2

Let w = 3x so cos(w) = √3/2

Using the unit circle, w = π/6 and w = 11π/6 but since we must divide these answers by 3, it is NOT SUFFICIENT to only consider these results. We must continue going around the unit circle so:

w = π/6

w = 11π/6

w = 13π/6

w = 23π/6

w = 25π/6

w = 35π/6

Now, back substitute and we get 6 answers:

3x = π/6 so x = π/18

3x = 11π/6 so x = 11π/18

3x = 13π/6 so x = 13π/18

3x = 23π/6 so x = 23π/18

3x = 25π/6 so x = 25π/18

3x = 35π/6 so x = 35π/18

Notice that 35π/18 is still less than 2π

Answer 3

4cos3x - 2√3 = 0; cos3x = √3/2

3x = π/6 + 2πn or 3x = 11π/6 + 2πn;

x = π/18 + 2πn/3 or x = 5π/18 + 2πn/3.

Now verying n = 0, 1, 2 we get all 6 solutions in interval [0, 2π);

n =0, x = π/18 or x = 5π/18

n = 1, x = 13π/18 or x = 17π/18

n = 2, x = 25π/18 or x = 29π/18

For n =3, 4, 5,.. we will have repetition

Hapyy New year, sorry!

Answer 4

4cos3x = 2sqr3

cos3x = (1/2)sqr3 = .5(1.732) = 0.866

3x = cos^-1(.866) = 30 degrees or pi/6 and also -30 degrees = 360-30 = 330 degrees or 11pi/6

x = 10 degrees or pi/18 radians

or 110 degrees or 11pi/18 radians