help with trig problem

[EmperorNero]

3tan^2*2x-1=0

one of the answers is pi/12. but how do I actually work it out step by step?

 

[Stealth1024]

This is similar to one of the many problems were are doing tonight as homework... lol

I would help you but I am busy working out mine. and its late... I'll be back tomorrow after school!

 

[Zoson]

eh?? as you wrote the question it does not make sense, are you sure you didn't mean...
3tan^2(2x-1)=0?
if you did it can be rewritten as
3[tan(2x-1)]^2=0
in which case you can just use regular trig identities to solve it
3(tan(2x-1))(tan(2x-1))=0
etc etc
-Zoson

 

[EmperorNero]

I'm not too positive, but I think your equation isn't right, zoson. I subbed in pi/12 in the original equation and the answer came out to be 0 which is what it's supposed to be. I then subbed pi/12 in your equation and came out with 1.



<< eh?? as you wrote the question it does not make sense, are you sure you didn't mean...
3tan^2(2x-1)=0? >>



yeah, 3tan^2(2x-1)= 0 is what I meant. sorry for the ambiguity.

 

[Zoson]

Hrmm, well I haven't done trig in awhile,
3tan^2(2x-1)=0
try to isolate x... are you allowed to use logarithms? You could try that.
otherwise maybe inverse trig functions...
or heck, just equivalents might work.
just try to solve for x? (but like me saying this is telling you something you didn't know lol)
If I still had my trig book I'm sure I could help more hehe.
-Zoson

 

[The Wildcard]

Damm, a trig book would be helpful, lol. Well i managed to simpfly it to

1 = cos^2(2x-1)

hehe, and i dunno how to go from there. I also don't know if that's right.

 

[Mday]

assuming you mean 3 (tan(2x-01))^2 = 0, the tangent of (2x-1) is squared...
--
yeah, wildcard, that is right
--
from the identity: tan^2 + 1 = sec^2... forgive the lack of (x) ;-)

from there you have what wildcard has:

cos^2 (2x-1) = 1

remember the cosine curve. cos^2(n) = 1, assume n = [0,2pi] which is the same as 0 <= n <= 2pi, which is the whole circle. <= is less than or equal to.

and when is cos^2(n) = 1? when when cos(n) = +-1 (plus or minus 1). also assuming you are dealing with the real numbers

and when is cos(n) = 1? when n = 0 or 2pi
and when is cos(n) = -1? when n = pi

so, you have 2x-1 = 0 OR 2x-1 = 2pi OR 2x-1 = pi. so x = 1/2, pi + 1/2, or (pi + 1)/2

where did you get pi/12?

--

if you use the other identity, sin^2 + cos^2 = 1, you get, sin^2(2x-1) = 0

when is sin^2(n) = 0? when sin(n) = 0, n = 0, pi, 2pi, which gives you the same answer as above.

--

or directly

3tan^2(2x-1) = 0

divide both sides by 3.

tan^2(2x-1) = 0.

so tan(2x-1) = 0

2x-1 = 0, pi, 2pi...

 

[The Wildcard]

Yeah those are the same answers i got, but i have no idea how he got pi/12 as an answer.

 

[EmperorNero]

the teacher gave us the answers so we can double check later...so that's where i got the answer.

 

[hendon]

i think he means
3 ( tan(2x) )^2 - 1 = 0
which becomes
( tan (2x) ) ^2 = 1/3
or tan 2x = +- 1/sqrt(3)
then 2x = pi/6, ...

so u get pi/12 as one of the answers...