find all solutionsa of the equation in interval [0,2pie)?

Question

3tan^3x=tanx
my solution:
tanx(3tan^2x-1)=0
tanx=0 > x = 0,pie (how do i get this PIE FROM)
or 3tan^2x -1=0
tanx=+-radical3/3
x = pie/6,5pie/6,7pie/6,11pie/6(HOW DO I GET ALL THESE PIES FROM?)
(I just don't undrestand how to get these pie/6,5pie/6,7pie/6,11pie/6 from?)

Answer 1

Your math is correct that the solutions are

tanx=0, √3/3, -√3/3

You don't seem to know what "pie" is though.

"pie" is represented by the greek symbol π (maybe you recognize this).

There are 360 degrees in a circle which can also be stated as there are 2π radians in a circle (think of the formula for the circumference of the circle = 2πr when the radius is 1).

The interval [0,2π) is just a circle.

To solve just figure out when tanx takes these values within the interval.

tanx=0 when x=0 and x=180 degrees. Converting this to radians gives x=0 and x=π (since 180 degrees is half a circle, it must be equal to π radians which is half of a circle as well since the circle is 2π radians).

tanx=√3/3 when x=30 degrees and x=210 degrees. Converting this to radians gives x=π/6 and x=7π/6

tanx=-√3/3 when x=150 degrees and x=330 degrees. Converting this to radians gives x=5π/6 and x=11π/6

So all the solutions are

x = 0, π/6, 5π/6, π, 7π/6, 11π/6

Answer 2

If you are talking only about quadrant I, there will be no negative values of tan x. The only zero value will be at x = 0 and the positive value will be at x = π/6. However, the interval involves more than just the one quadrant. Because we are talking about the tangent, the identity
tan x = tan (π + x) comes into play from quadrant III. Therefore, if x is a solution, then so is x + π. Because x = 0 is a solution, so is x = π. Because x = π/6 is a solution, so is x = π + π/6 or 7π/6.

But this isn't all. We also have the identities -tan (x) = tan (-x) and tan (x) = tan (x + 2π). Because tan x can also be -√3/3 = -tan (π/6) = -tan(7π/6), we would pick up the solutions x = -π/6 and -7π/6 if they were in our domain interval. However, if we add 2π to each of these, we get another two solutions, 11π/6 and 5π/6, that are in the domain interval, in quadrants IV and II.

Answer 3

I think there's something misleading, else I'm very wrong.

A closed interval uses brackets: [0, pi] or 0= < x = < pi
An open interval uses parentheses: (0, pi) or 0 < x < pi
A half open interval (or half closed interval) doesn't include one (of 2 points) [0, pi) is the same as 0= < x < pi

Now to your other problem, the RELATED angle for
a particular trig function is the pos acute angle whose trig function = absolute value of the trig function being considered.

For example in the interval [pi/2, 3pi/2] there is only one q whose sin q = 1/2 and only one w whose sin w = - 1/2 BUT
both q and w have the related angle of 30 degrees associated with them. Once you determine the related angle then your task becomes to put that related angle into the quadrant(s) of your universe for solution(s).

So when your tan x = (+ or -) sqrt3/3 (or {+ or - }1/(sqrt3) ) you should be thinking related angle of 30 degrees, then since your universe for solution(s) is [0, 2pi) you end up making that 30 angle with the x axis in all 4 quadrants.

Answer 4

interval [0, 2pi) means 0 is not included but 2pi is included

3tan^3(x) = tan(x)

3tan^3(x) - tan(x) = 0

tan(x)[3tan^2(x) - 1] = 0

either tan(x) = 0 or

3tan^2(x) = 1

when tan(x) = 0

x = pi, 2pi (since 0 is not included)

when 3tan^2(x) = 1

tan^2(x) = 1/3

tan(x) = ± 1/√3

x = pi/6 (1st quadrant) (+)

x = pi + (pi/6) = 7pi/6 (3rd quadrant)(+)

x = pi - (pi/6) = 5pi/6 (2nd quadrant)(-)

x = 2pi - (pi/6) = 11pi/6 (4th quadrant)(-)

Answer 5

when u solve the problem,x=0 is your solution,but tanx is also zero at x=pi,u can find that by graphing y+tanx on your calculator and find out at what points the graph intersects the x-axis,you will find another point at x=pi.
in the second solution you got tanx=+-radical3/3,do the same , graph y=tanx,y=radical3/3 and y= - radical3/3find all points of intersection of tanx, with the two lines.

Answer 6

3tan^3x=tanx
3tan^2x =1
tan^2x = 1/3
tan x = +/-sqrt(1/3) = +/- sqrt(3)/3
x = arctan +/- (sqrt(3)/3)