Question

tan(2Θ)+tan(Θ)= 0
SOLVE ON THE INTERVAL: 0≤Θ<2π

Answer 1

@mustafa2014 help pls?

Answer 2

\[\tan(2Θ)+\tan(Θ)= 0\]
\[\frac{ 2\tan(Θ) }{ 1-\tan^2(Θ) }+\tan(Θ)=0\]
\[\frac{ 2\tan(Θ) }{ 1-\tan^2(Θ) }+\frac{ tan(Θ)(1-\tan^2(Θ) }{ 1-\tan^2(Θ) }=0\]
\[\frac{ 2\tan(Θ)+tan(Θ)-tan^3(Θ) }{ 1-\tan^2(Θ) }=0\]
\[3\tan(Θ)-tan^3(Θ)=0\]

Answer 3

wait so after 3tanΘ and all that do u factor

Answer 4

\[\tan(Θ)(3-tan^2(Θ)) =0\]

Answer 5

and then u solve for each individually right?

Answer 6

like make both= 0 and solve?

Answer 7

yep

Answer 8

is that the right way though? for sure?

Answer 9

thank you again.

Answer 10

wut

Answer 11

wait no thats not how youd solve. thats only if theyre muliplying each other

Answer 12

@Mertsj does this seem right ? it seems wrong somehow?

Answer 13

Is the original problem tan(2x) or tan^2x

Answer 14

its tan 2theta not squared

Answer 15

\[3\tan \theta -\tan ^3\theta =0\]
\[\tan \theta (3-\tan ^2\theta)=0\]
\[\tan \theta=0; or\]
\[3-\tan ^2\theta=0\]

Answer 16

Solve each equation

Answer 17

where r u getting the 3 from??

Answer 18

the original equaion is tan(2Θ)+tan(Θ)= 0

Answer 19

Did you not follow the work that mustafa posted above?

Answer 20

He substituted for tan 2theta and simplified for you.
He even did the factoring for you.
All you had to do was set each factor equal to 0 and solve
So I set each factor equal to 0
Now all you have to do is solve.

Answer 21

If you do not trust his work (and I did not check it) all you have to do is replace tan 2theta with 2tan(theta)/1-tan^2(theta) and simplify for yourself.

Answer 22

yes but someone else helped me w/ same problem and the final answer they got was Θ= sqrt 3 and then you got the solutions.

Answer 23

theta is an angle. If they got tan theta = sqrt 3, that would make sense because then theta would be 60 degree.
But theta = sqrt 3 is suspect.

Answer 24

o=sorry it was tanΘ=sqrt 3 so is that the same thing as what mustafa said?

Answer 25

wait it is right? they both said the same thing then?