Solve for value

**taurus** · September 4th 2007, 02:14 AM

Find the exact values of x between 0 and 360 for

2 cos^3(x) - cos(x) = 0

-----------------------------------

So what i did was cos(x) = x

therefore 2x^3 - x = 0

Now i got stuck there??

**ThePerfectHacker** · September 4th 2007, 03:51 AM

Originally Posted by taurus

Find the exact values of x between 0 and 360 for

2 cos^3(x) - cos(x) = 0

-----------------------------------

So what i did was cos(x) = x

therefore 2x^3 - x = 0

Now i got stuck there??

$\cos x(2\cos^2 x - 1)=0$
Use identity,
$\cos x \cos 2x = 0$
Thus,
$\cos x = 0 \mbox{ or }\cos 2x = 0$ .
Now continue ...

**taurus** · September 4th 2007, 04:39 AM

Originally Posted by ThePerfectHacker

$\cos x(2\cos^2 x - 1)=0$
Use identity,
$\cos x \cos 2x = 0$
Thus,
$\cos x = 0 \mbox{ or }\cos 2x = 0$ .
Now continue ...

still a bit confused. so i use identity sin(2x) = cos(2x)?

**topsquark** · September 4th 2007, 04:47 AM

Originally Posted by taurus

Find the exact values of x between 0 and 360 for

2 cos^3(x) - cos(x) = 0

The trig identity approach is confusing you. And please refer again to ThePerfectHacker's post: it is not $sin(2x) = cos(2x)$ . This is not an identity! The identity he used is $cos(2x) = 2cos^2(x) - 1$ .

Let's try it this way:
$2cos^3(x) - cos(x) = 0$ <-- Factor the common $cos(x)$

$cos(x)(2cos^2(x) - 1) = 0$

Now either
$cos(x) = 0 \implies x = 90^o, 270^o$
or
$2cos^2(x) - 1 = 0 \implies cos(x) = \pm \sqrt{ \frac{1}{2}} \implies x = 45^o, 135^o, 225^o, 315^o$

-Dan

**taurus** · September 4th 2007, 05:07 AM

Originally Posted by topsquark

The trig identity approach is confusing you. And please refer again to ThePerfectHacker's post: it is not $sin(2x) = cos(2x)$ . This is not an identity! The identity he used is $cos(2x) = 2cos^2(x) - 1$ .

Let's try it this way:
$2cos^3(x) - cos(x) = 0$ <-- Factor the common $cos(x)$

$cos(x)(2cos^2(x) - 1) = 0$

Now either
$cos(x) = 0 \implies x = 90^o, 180^o$
or
$2cos^2(x) - 1 = 0 \implies cos(x) = \pm \sqrt{ \frac{1}{2}} \implies x = 45^o, 135^o, 225^o, 315^o$

-Dan

im still not quite getting it, ur saying cos(2x) but i dotn have cos(2x) in my expression? ummmm sorry
you lost me by the factoring?

also here:
$cos(x)(2cos^2(x) - 1) = 0$

whats happening there?

thanks

**janvdl** · September 4th 2007, 06:50 AM

$2cos^3 (x) - cos (x) = 0$

Set $cos(x) = k$

$2k^3 - k = 0$

$k(2k^2 - 1) = 0$

Now set $k = cos(x)$ like it was

$cos x ((2cos^2 x) - (1)) = 0$

But we know that $2cos^2 (x) - 1 = cos 2x$ from our identities

So then $(cos x)(cos 2x) = 0$

For $Cos x = 0$

$x = Cos^{-1} \ 0$

$x = 90 \ or \ x = 270$

For $Cos 2x = 0$

$2x = Cos^{-1} \ 0$

$x = \frac{Cos^{-1} \ 0}{2}$

$x = 45 \ or \ x = 135 \ or \ x = 225 \ or \ x = 315$

**taurus** · September 4th 2007, 07:14 AM

ok right, so u jus solve for each of those cos

**janvdl** · September 4th 2007, 07:16 AM

Originally Posted by taurus

ok right, so u jus solve for each of those cos

Yes

Oh and do refrain from using chat language, it's against the rules and you'll get in trouble.

**taurus** · September 4th 2007, 07:18 AM

wait ur answers r different to that other guys?

**janvdl** · September 4th 2007, 07:21 AM

Originally Posted by taurus

wait ur answers r different to that other guys?

He made a mistake saying $cos(180) = 0$
I'll check his others now.

EDIT: His other solutions are correct.

EDIT(2): Sorry i only forgot to write 225 and 315 as solutions. You will notice that $Cos(2x) = 0$ is equal to 4 answers.

**topsquark** · September 4th 2007, 11:34 AM

Originally Posted by janvdl

He made a mistake saying $cos(180) = 0$

Whoopsie! I fixed it, thanks for the spot.

-Dan

**janvdl** · September 5th 2007, 03:06 AM

Originally Posted by topsquark

Whoopsie! I fixed it, thanks for the spot.

-Dan

You do realise Topsquark that I have printed that one out and framed it against my wall...

**topsquark** · September 5th 2007, 05:31 AM

Originally Posted by janvdl

You do realise Topsquark that I have printed that one out and framed it against my wall...

Well because it was you that corrected me I seriously thought about coming up with a really complicated argument involving Green's functions

and doing a lot of chest thumping, but I decided not to.

-Dan

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