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.
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