Question

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

7 tan3x - 21 tan x = 0

Answer 1

Is that \(\large\rm \tan(3x)\) or \(\large\rm \tan^3x\) ? :)

Answer 2

tan ^3 x

Answer 3

\[\large\rm 7\tan^3x-21\tan x=0\]Let's factor some stuff first.
Looks like they both have..... a tangent... and a 7, ya?

Answer 4

7tanx(tan^2x-3)=0

Answer 5

Ok good!
Now we apply our `Zero-Factor Property`, setting each individual factor equal to zero, and then solving from there.

So our first factor equal to zero is:
\(\large\rm 7\tan x=0\)

Answer 6

You can divide by 7 to simplify things down a bit:
\(\large\rm \tan x=0\)
Understand how to solve for x in this case? :)

Answer 7

would you plug it into a calculator using tan^-1?

Answer 8

I guess you `could` do that :p
it's better just remember some of your special angles.

tangent is 0 when the angle x is 0.

Answer 9

Oh but I guess umm... they want all of the solutions from 0 to 2pi, so that produces another angle as well.

Answer 10

oh yeahhhh, so then pi and 0? or would it be 1/pi and 3/pi?

Answer 11

sorry got distracted :)
um um... yaaaaa, 0 and pi sound good for our first two solutions.

Answer 12

Applying the zero-factor property again:\[\large\rm \tan^2x-3=0\]

Answer 13

So ummm... how bout.. add 3.
then square root or something, ya?

Answer 14

\[\large\rm \tan^2=3\]

Answer 15

\[\large\rm \tan x=\pm\sqrt{3}\]

Answer 16

The positive root 3 will give you 2 angles again.
while the negative root 3 will give you 2 more angles!

So this solution set is actually giving us 4 angles! kinda crazy.

Answer 17

uhhhhh... how do you get your solutions from \[\pm \sqrt{3}\]

Answer 18

Hmm you gotta get better with your unit circle missy! :O
http://4.bp.blogspot.com/-FMVojhlkcSQ/UInlIKO88oI/AAAAAAAAAC4/-fzMOz6di2Y/s1600/image010.jpg
So umm.. i couldn't find a really good picture, but here is one.

Answer 19

The coordinate pair lists first the sine value of that angle, and then the cosine.
And then outside of the brackets, on the right, is the tangent value of that angle.

So if we look in the first quadrant, it looks like the angle pi/3 gives us sqrt(3), ya?

Answer 20

Ohhhhh! I see now! :P Thank you so much!

Answer 21

cool c: