SOLUTION: Find all the solutions in the interval [0,2pi) in radian measure 3tan^2 2x-1=0

Algebra.Com
Question 1010969: Find all the solutions in the interval [0,2pi) in radian measure
3tan^2 2x-1=0

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
tan^2(x) = 0 when x ={0, pi, 2pi} as shown below:

$$$

tan^2(2x) = 0 when 2x = {0, pi, 2pi}.
solve for x and you get x = {0, pi/2, pi, 3pi/2, 2pi} as shown below:

$$$

tan^2(2x-1) = 0 when (2x-1) = {0, pi, 2pi}
solve for x and you get x = {.5, 2.071, 3.642, 5.212} as shown below:

$$$

having access to a grahping calculator so you can see what's going on helps a lot.

the calculator i used is at http://www.desmos.com/calculator.

how would you solve this algebraically?

see below for what i did.

you start with 3tan^2(2x-1) = 0

divide both sides of this equation by 3 to get:

tan^2(2x-1) = 0

take the square root of both sides of this equation to get:

tan(2x-1) = 0

this is true if and only if tan^-1(0) = 2x-1

you know that tan is 0 when the angle = 0 or 180 or 360.

in radians, tan is 0 when the angle = 0 or pi or 2pi.

since the angle we are looking for is represented by (2x-1), this means that (2x-1) = 0 or pi or 2pi.

solve for x to get x = .5 or 2.070796327 or 3.641592654

since the formula is tan^2(2x-1) = 0, the frequency is 2 which means the period is pi instead of 2pi.

in order to get results up to 2pi, you need to go through 2 periods.

this means the values of x will repeat every pi radians.

.5 + pi = 3.641592654 + pi = 6.783185307.

2.070796327 + pi = 5.212388981 + pi = 8.35981634

since 2pi = 6.283185307 radians, you only need the values that are between 0 and 6.283185307 radians.

those values are:

.5, 2.070796327, 3.641592654, or 5.212388981 radians.

those are the values shown on the final graph rounded to 3 decimal places.

you should see the following values on the third graph abo ve.

.5, 2.071, 3.642, 5.212.

to test if these values for x are correct, take one of them and follow the original formula to see if it rsults in 0.

for example, tan^2(2x-1) = 0 becomes tan^2(2*3.641592654 - 1) = 0 when x = 3.641592654.

i used my calculator to get tan^2(2*3.641592654-1) = 6.7305616 * 10^-19 which is extremely close to 0.

3.64592654 is a rounded number.

the internally stored number has more decimal places than that.

since 3.6459... is the result of .5 + pi, i'll use that instead to get tan^2(2*(.5+pi)-1) = 0

the difference is in the rounding.













RELATED QUESTIONS

Find all solutions in the interval [0, 2pi].... (answered by lwsshak3)
find all solutions: 3sin^2x+2sinx-1=0; in the interval... (answered by ikleyn)
Find all radian values of theta in the interval 0 (answered by Fombitz)
Find all solutions of the equation in the interval [0, 2pi).... (answered by Alan3354)
could you help me find all the solutions in the interval [0, 2pi)... (answered by lwsshak3)
Find all solutions in the interval o and 2pi 1.) Cos^2x+ sinx +1= 0 (answered by lwsshak3)
Find all solutions of each equation in the interval [0, 2pi)... (answered by lwsshak3)
Find all solutions of each equation in the interval [0, 2pi)... (answered by lwsshak3)
Find all solutions of each of the equations in the interval (0, 2pi) cos(x- pi/2) +... (answered by lwsshak3)