Question

3tan^3x=tanx

Answer 1

ok - try this yourself first using the same technique as last time

Answer 2

\[3\tan^3x=tanx\]

Answer 3

\[3\tan^3x-tanx=0\]

Answer 4

good - now factorise

Answer 5

\[\tan(3\tan^2x-1)=0\]

Answer 6

yup - small omission of 'x' in tan(x) outside the braces

Answer 7

Oh, woops, keyboard lag

Answer 8

:)

Answer 9

\[tanx=0, x=0\]

Answer 10

so here you need to use the zero product property - do you know this rule?

Answer 11

i.e. if:\[a\times b=0\]then either a=0 or b=0

Answer 12

\[3\tan^2x-1=0 \tan^2x=1/3\]

Answer 13

Its more like \[\tan(x) = 0 \]

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

Answer 14

@LordVictorius - yes you got that right

Answer 15

\[tanx=\sqrt(1/3)\]

Answer 16

Where would I go from here?

Answer 17

do not forget the +/- for the square root

Answer 18

Oh ya.

Answer 19

and also:\[\tan(x)=\pm\sqrt{\frac{1}{3}}=\pm\frac{\sqrt{1}}{\sqrt{3}}=\pm\frac{1}{\sqrt{3}}\]

Answer 20

What's the difference?

Answer 21

Just simplifying?

Answer 22

so your solutions are:\[x=\tan^{-1}(0)\]and:\[x=\tan^{-1}(\pm\frac{1}{\sqrt{3}})\]

Answer 23

The teacher said I need them in radians.

Answer 24

yes, so you need to find the angles that give to a tan of zero for the first one

Answer 25

which is not just x=0

Answer 26

e.g \(\tan(\pi)=0\)

Answer 27

pi/6

Answer 28

so for \(\tan(x)=0\) we have \(x=n\pi\) for \(n=0,1,...\)

Answer 29

Okay, I'm following you so far.

Answer 30

you need a similar /general/ solution for \(\tan(x)=\pm\frac{1}{\sqrt{3}}\)

Answer 31

pi/6 is just one particular solution

Answer 32

Hmm.

Answer 33

I don't get it.

Answer 34

do you know this rule:\[\tan(\theta)=\tan(\theta+\pi)\]

Answer 35

Now I do.

Answer 36

each trig function has a periodicity

Answer 37

you might find this useful: http://www.clarku.edu/~djoyce/trig/identities.html

Answer 38

What anaseer is saying that look at the unit circle..