Ask your own question, for FREE!
Mathematics 52 Online
OpenStudy (anonymous):

how to solve equation 3tan^2x - 2tanx = 0 i wrote the steps so far but i dont know next http://bamboodock.wacom.com/doodler/a1734a20-e823-4a64-9bff-80ea08d8af33

OpenStudy (anonymous):

Find tan^{-1}(2/3) now..

OpenStudy (anonymous):

Or you want to find general solution for this equation ??

OpenStudy (anonymous):

i tried but thers so many numbers coming out

OpenStudy (anonymous):

Like ..

OpenStudy (anonymous):

33.69006753

OpenStudy (anonymous):

This is one number only..

OpenStudy (anonymous):

So there will be two values of x: 0 and 33.7..

OpenStudy (anonymous):

why 0?

OpenStudy (anonymous):

from first one..

OpenStudy (anonymous):

tanx = 0 implies x = 0..

OpenStudy (anonymous):

im lost

OpenStudy (anonymous):

Do you have final answers with you for this one ??

OpenStudy (alexwee123):

you can always treat it as a quadratic to....

OpenStudy (alexwee123):

too*

OpenStudy (anonymous):

yeah its 0, 33.7˚, 180˚, 213.7˚

OpenStudy (anonymous):

wait so tan^1(2/3) was 33.69.... i round it off to 33.7˚!!!!

OpenStudy (anonymous):

@lsugano this is called finding the general solution that is what I was asking to you..

OpenStudy (anonymous):

oh im sorry!! i ddnt know what general solution meant :((((

OpenStudy (anonymous):

For tanx = 0: the general solution will be: \[\large x = n \pi, n \in \mathbb {Z}\]

OpenStudy (anonymous):

General solution means to find all the values of x in any trigonometric equation..

OpenStudy (anonymous):

For \(tanx = tany\) General solution is: \[\large x = n \pi + y\]

OpenStudy (anonymous):

Here : as y = tan^{-1}(2/3) so general solution of second equation becomes: \[\large x = n \pi + \tan^{-1}(2/3)\]

OpenStudy (alexwee123):

|dw:1343980574390:dw| take the inverse tan of each and your answers are |dw:1343980628234:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!