1. Solve for value

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??

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

3. ?

Originally Posted by ThePerfectHacker
$\displaystyle \cos x(2\cos^2 x - 1)=0$
Use identity,
$\displaystyle \cos x \cos 2x = 0$
Thus,
$\displaystyle \cos x = 0 \mbox{ or }\cos 2x = 0$.
Now continue ...
still a bit confused. so i use identity sin(2x) = cos(2x)?

4. 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 $\displaystyle sin(2x) = cos(2x)$. This is not an identity! The identity he used is $\displaystyle cos(2x) = 2cos^2(x) - 1$.

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

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

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

-Dan

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

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

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

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

-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:
$\displaystyle cos(x)(2cos^2(x) - 1) = 0$

whats happening there?

thanks

6. $\displaystyle 2cos^3 (x) - cos (x) = 0$

Set $\displaystyle cos(x) = k$

$\displaystyle 2k^3 - k = 0$

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

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

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

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

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

For $\displaystyle Cos x = 0$

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

$\displaystyle x = 90 \ or \ x = 270$

For $\displaystyle Cos 2x = 0$

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

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

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

7. thanks

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

8. 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.

9. wait ur answers r different to that other guys?

10. Originally Posted by taurus
wait ur answers r different to that other guys?
He made a mistake saying $\displaystyle cos(180) = 0$
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 $\displaystyle Cos(2x) = 0$ is equal to 4 answers.

11. Originally Posted by janvdl
He made a mistake saying $\displaystyle cos(180) = 0$
Whoopsie! I fixed it, thanks for the spot.

-Dan

12. 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...

13. 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