trigonometric equation

[spice trout]

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[wet crane]

Try to divide things that apeer on both sides. if you need more hints let me know

[cold spire]

But also note that if you do so, you require that the things you divide not equal zero.

[wet crane]

No

[cold spire]

...by doing what?

[wet crane]

I know but you worry about that later

$4\sin^2 x \tan x = 3\tan x$

[warped skiff]

TheVinkler

[wet crane]

First you see that those 2 tans? you can divide by tanx as long as tanx not equal to 0

At first yes

$4\sin^2 x = 3$

[warped skiff]

TheVinkler

[wet crane]

You probably want to isolite for x right? how do you think we should start?

No you divide by tan

Actually now that I thought about it there is an easier solution

Yes

Would you like to see the easier solution?

Ok so you subtract 3tan x

$4\sin^2 x \tan x = 3\tan x$

[warped skiff]

TheVinkler

[wet crane]

$4\sin^2 x \tan x -3\tan x= 0$

[warped skiff]

TheVinkler

[wet crane]

Now you can factor tan x right?

You know factoring right?

Let $\tan x = u$

[warped skiff]

TheVinkler

[wet crane]

Ok?

Just wait a sec ok?

You will see

$4\sin^2 x \tan x -3\tan x= 0\to 4\sin^2(x) \cdot u-3u$

[warped skiff]

TheVinkler

[wet crane]

Can you see now how we can factor?

IGNORE THE SIN!

Can you see how to factor

Ok you see that both are bgin multplied by u right?

Just for me to help you better, what grade are you? or is this like algebra 1 or algebra 2

Just to know the depth

Becuse you are foucosing on losing the sin when you need to do things before hand

The sin makes you nervous so you try to get rid of it

What you should always do in thsese senrios is look for something in common between the sides

When you see this:

$4\sin^2 x \tan x = 3\tan x$

[warped skiff]

TheVinkler

[wet crane]

I want you to see this:

$4\sin^2 x {\color{red}\tan x} = 3{\color{red}\tan x}$

[warped skiff]

TheVinkler

[wet crane]

Ok?

Ok

Factor the tanx from both sides

$\tan x (4\sin^2x-3)=0$

[warped skiff]

TheVinkler

[wet crane]

There is

Ok thing about this

If I have

$ax+bx$

[warped skiff]

TheVinkler

[wet crane]

I can also write it as

$x(a+b)$

[warped skiff]

TheVinkler

[ornate gazelle]

reverse distributive property

or factoring

whatever term youve heard more

[ornate gazelle]

pay attention to these two things vinkler told you about

[ornate gazelle]

.

hmm

https://www.youtube.com/watch?v=zRgnVbh6psI
see if this helps

or we could just go back to dividing tanx from both sides if that makes more sense

[wet crane]

No

[ornate gazelle]

3tanx MINUS tanx is 2tanx

you treat tanx the same way youd treat a variable like u

u=3u
when you divide by u,
1=3

[wet crane]

Yes

You can do that

You can do both, diffrent letters though

you can say that sinx = u and tanx = v

Yes

and then use your solution of u to solve for x

Great!

What is U equal to?

Yes but what did we subsitute u with?

sin x

thats right so we get

YES!

LETS GO

Inverse sin

yes

what did you get?

Yes'

Depends how you want your answers

yes

Just use it in degrees

yes but this $\frac{\pi}{3}$

[warped skiff]

TheVinkler

[wet crane]

YES!

LETS GO

WE WINDOWS

now there is also one more solution

Yep

There are actaully inifinte solutions since

$\sin({x+2\pi})=\sin x$

[warped skiff]

TheVinkler

[wet crane]

No

because of the tan

But another solution is 0

$\tan 0 = 0$

[warped skiff]

TheVinkler

[wet crane]

.

Then you assumed that $x\ne0$

[warped skiff]

TheVinkler

[wet crane]

But if x=0 then this also works

One set of solutions is

$x=2\pi n$ and another $x=2\pi n + \frac{\pi}{3}$

[warped skiff]

TheVinkler

[wet crane]

Our solution were $x=0$ and $x=\frac{\pi}{3}$

[warped skiff]

TheVinkler

[wet crane]

right?

So we add $2\pi n$

[warped skiff]

TheVinkler

[wet crane]

because $\tan(x)=\tan(x+2\pi n)$

when n is a natrual number

[warped skiff]

TheVinkler

[wet crane]

these are peridioc functions

These are the rest, for all natrual numbers n

So just write 0 and pi/3

I don't seem to understand

OG

MB

there are finite solutions since they said that x is within $[0,2\pi]$

[warped skiff]

TheVinkler

[wet crane]

So ye you get $|\sin(x)|=\frac{\sqrt{3}}{2}$ which is $\pi/3$ and $5/pi/3$

[warped skiff]

TheVinkler

[wet crane]

0 Is still a solution

Yes

[wet crane]

Also show (by caluclation) that if x=0 it also works

what?

no

$4\sin^2 x \tan x = 3\tan x$

[warped skiff]

TheVinkler

[wet crane]

This was the original right?

so put tanx = 0

Yes

it was

Oh

mb

Mb

Just a sec

This right?

$3sin^2x+2=7$

[warped skiff]

TheVinkler

[wet crane]

Yes if you got sinx =2 there is no solution

sinx = 1/3 is okay

because $-1\le \sin x \le 1$

[warped skiff]

TheVinkler

[wet crane]

always

If sinx>1 that means there is a right triangle in which the hypothenes isn't the largest side

$\sin^2x=\frac{5}{3}$

[warped skiff]

TheVinkler

[wet crane]

Doen't have any solutions

[ornate gazelle]

the question