Question

find value of x in the interval [0,2π/3] such that 2cos(3x)-1=0 .

I got up to:

cos(3x) =1/2
but after that I am lost :(

Answer 1

so you know for which x you get
\[\cos(x)=\frac{1}{2}\]?

Answer 2

i believe on value sis
\[\frac{\pi}{3}\]
so that means
\[3x=\frac{\pi}{3}\] solve to get
\[x=\frac{\pi}{9}\] as one answer

Answer 3

What do you mean?

Answer 4

first you need a value that gives
\[\cos(x)=\frac{1}{2}\] right?

Answer 5

then replace x by 3x and solve. if you know that
\[\cos(\frac{\pi}{3})=\frac{1}{2}\] that means
\[3x=\frac{\pi}{3}\] and so dividing by 3 to get x gives
\[x=\frac{\pi}{9}\]

Answer 6

2cos(3x) - 1 = 0
2cos(3x)=1
cos(3x) = 1/2
3x = acos(1/2)
x = acos(1/2) / 3
= pi/3/3
= pi/6

Ans pi/6

Answer 7

not quite, but close. last step was not right.

Answer 8

i.e pi/9

Answer 9

look at the unit circle on the last page of the cheat sheet. find the places on the unit circle where the first coordinate is 1/2
then look at the angle. those will be the angles with cosine of 1/2

Answer 10

Thanks @satelite73 soooooo much! You saved my life. I have another question if you don't mind