Question

tan 2x- cot x=0 in the iinterval 0,2pi

Answer 1

Well no matter what, you have to reduce tan(2x) down to a single angle, not 2 like you have. Happen to have the double angle formula for tangent handy on ya?

Answer 2

yes,
2 tan u/1-tan^2 u

Answer 3

Alright, awesome. So since doing that identity gives ua a bunch of tangents, I would suggest turning your cotangent into a tangent as well.

Answer 4

okay, still not seeing a way to finish it tho

Answer 5

We'll get there :P So we can turn cot into tan by saying cotx = 1/tanx, so we would then have this so far

\[\frac{ 2tanx }{ 1-\tan^{2}x }- \frac{1}{tanx}\]From here I would combine this into one fraction by multiplying top and bottom of the left fraction by tanx and by multiplying top and bottom of the right fraction by 1-tan^2(x), giving us:
\[\frac{ 2tanx(tanx)-1(1-\tan^{2}x) }{ tanx(1-\tan^{2}x) } \implies \frac{ 3\tan^{2}x - 1 }{ tanx(1-\tan^{2}x)}= 0\]
Now from here we need to realize that solving the denominator does us no good. Why? Well, if we multiplied both sides by the denominator, it would disappear. If we set the denominator equal to 0, we'd only get undefined values. Basically at this point, our denominator means nothing. So we just want to solve 3tan^2(x) - 1 = 0

Answer 6

so essentially where does tan equal 1/3?

Answer 7

Where does tan^2(x) = 1/3. We still have the 2nd power there to deal with.

Answer 8

ya thats what i was thinking lol

Answer 9

\[tanx = \pm \sqrt{\frac{1}{3}}\implies tanx = \pm \frac{ \sqrt{3} }{ 3 }\]

Answer 10

hmph

Answer 11

what gave u the clues to do that

Answer 12

To do what exactly?

Answer 13

i guess to make it into one fraction, but it only makes sense to make it into one fraction.

Answer 14

cause you are still just subtracting them

Answer 15

Not only that, but in my experience doing that usually just turns out to be the best the best thing to do. Especially since we have a bunch of tangents, you put them into one fraction and see what can simplify. I also know that when I do so, whatever is in the denominator becomes worthless to solving the problem.

Answer 16

okay makes sense

Answer 17

okay here is another question, just to rewrite the expression (cos x+sinx)(cos x -sinx) which would factor into sin^2 x -cos^2 x

Answer 18

rewrite it with double angle formulas

Answer 19

cos^2(x) -sin^(2)x

Answer 20

Which then yeah, goes to simply cos2x

Answer 21

why did i not see that?

Answer 22

Dunno, lol. But yeah, even looking at the foil, neither cosine is negative, so no way cos^2 would ever be negative.

Answer 23

okay

Answer 24

so if you have to use the power reducing formulas on cos^4 2x could you just do the
1+cos 2x/4?

Answer 25

Nah. Youd have to say this:
\[\cos^{4}2x = (\cos^{2}2x)^{2}= (\frac{ 1+\cos2x }{ 2 })^{2}\]

Answer 26

Whoops, correction.

Answer 27

okay

Answer 28

\[(\frac{ 1+\cos4x }{ 2 })^{2}\]

Answer 29

thank you so much you have been a GREAT help

Answer 30

Yeah, np ^_^

Answer 31

Wish u were at my school

Answer 32

They dont have any form of tutorial services at your school?

Answer 33

im sure they do, but i went to the teacher n thought i had it so i didnt look into it, and i was doing my homework and of course, he didnt explain well enough, so i was looking for all sources possible

Answer 34

at midnight you know how it is

Answer 35

Yeah. And this is a good place to come to when you cant get any other help. Yeah, either because its midnight or you just want to post a question before you go to bed/school/work and see if someone asnwered it later, anything.

Answer 36

very good place i love it. i learn more here than i do from teachers.

Answer 37

A lot of place and people who do tutoring or help on the side are better than teachers, sadly enough x_X

Answer 38

true to a lot of things.

Answer 39

what are those points on the x axis
y axis they are 1 0 -1 0 1