Find the exact values of x between 0 and 360 for
2 cos^3(x) - cos(x) = 0
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So what i did was cos(x) = x
therefore 2x^3 - x = 0
Now i got stuck there??
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
$\displaystyle 2cos^3 (x) - cos (x) = 0$
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 $