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??

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- Sep 4th 2007, 02:14 AMtaurusSolve 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?? - Sep 4th 2007, 03:51 AMThePerfectHacker
- Sep 4th 2007, 04:39 AMtaurus?
- Sep 4th 2007, 04:47 AMtopsquark
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 - Sep 4th 2007, 05:07 AMtaurus
- Sep 4th 2007, 06:50 AMjanvdl
$\displaystyle 2cos^3 (x) - cos (x) = 0$

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 $ - Sep 4th 2007, 07:14 AMtaurusthanks
ok right, so u jus solve for each of those cos

- Sep 4th 2007, 07:16 AMjanvdl
- Sep 4th 2007, 07:18 AMtaurus
wait ur answers r different to that other guys?

- Sep 4th 2007, 07:21 AMjanvdl
- Sep 4th 2007, 11:34 AMtopsquark
- Sep 5th 2007, 03:06 AMjanvdl
- Sep 5th 2007, 05:31 AMtopsquark