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# Pre-Calculus

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cos2x - cosx = 0 i need steps....

Guest Mar 3, 2016

#8
+20151
+10

cos2x - cosx = 0 i need steps....

rearrange

$$\begin{array}{rcll} \cos{(2x)} - \cos{(x)} &=& 0 \\ \cos{(2x)} &=& \cos{(x)} \\\\ \boxed{~ \cos{(\varphi)} = \cos{(-\varphi)}~}\\\\ \end{array}$$

We have 4 variations: {nl} $$\begin{array}{lrclrcl} (1) & \cos{(2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (2) & \cos{(-2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (3) & \cos{(2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\ (4) & \cos{(-2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\\\ (1) & 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\qquad \Rightarrow &\qquad x &=& 0 \pm 2k\pi\\ (2) & -2x \pm 2k_1\pi &=& x \pm 2k_2\pi \qquad \Rightarrow &\qquad -3x &=& 0 \pm 2k\pi\\ (3) & 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad 3x &=& 0 \pm 2k\pi\\ (4) & -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad -x &=& 0 \pm 2k\pi\\\\ & \Rightarrow x &=& \pm 2k\pi \\ & \Rightarrow x &=& \pm \frac{2}{3}k\pi \\ \end{array}$$

We can put together  $$x = \pm \frac{2}{3}k\pi \qquad k \in N_0$$

heureka  Mar 3, 2016
edited by heureka  Mar 3, 2016
edited by heureka  Mar 3, 2016
#1
+131
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x would 0.

= cos(2*0) - cos(0)

= 1 - 1

= 0

MATHBITCH  Mar 3, 2016
#2
+93866
+10

Hi Guest.

cos2x - cosx = 0

$$cos2x - cosx = 0\\ cos^2x-sin^2x-cosx=0\\ cos^2x-(1-cos^2x)-cosx=0\\ cos^2x-1+cos^2x-cosx=0\\ 2cos^2x-cosx-1=0\\ let \;\;y=cosx\\ 2y^2-y-1=0\\ 2y^2-2y+y-1=0\\ 2y(y-1)+1(y-1)=0\\ (2y+1)(y-1)=0\\ 2y=-1\;\;\;\;or\;\;\;\;y=1\\ y=-\frac{1}{2}\;\;\;\;or\;\;\;\;y=1$$

$$cosx=-\frac{1}{2} \qquad or \qquad cosx=1\\ for\;\; 0\le x<360\; degrees\\ x=180\pm 60 \;degrees \qquad or \qquad x=0\; degrees\\ General \;solutions\\ x=180\pm 60 \;+360n\;degrees \qquad or \qquad x=0+360n\; degrees\\ x=180\pm 60 \;+360n\;degrees \qquad or \qquad x=360n\; degrees\\ x=180(2n+1)\;\pm\ 60\;degrees \qquad or \qquad x=360n\; degrees\qquad n\in Z\\ \mbox{Normally this would be expressedin radians}\\ x=(2n+1)\pi\pm\frac{\pi}{3} \qquad or \qquad x=2\pi n \qquad \qquad n\in Z$$

Now that I have finished it has occurred tyo ma that maybe you wanted  the solutions to

cos^2x - cosx = 0

that is

(cosx)^2 - cosx = 0

which is a very much easier question ://

Melody  Mar 3, 2016
#3
+5252
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90% of all the math there is beyond me, at least, I think it is.

rarinstraw1195  Mar 3, 2016
#5
+93866
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I expect you are right rarinstraw but that is the beauty of this place.

Almost all of us see mathematics here that is new or unknown to them.  I regularly see things that Alan, Heureka, CPhill, Bertie and Geno3141 put up on the forum that are either totally beyond me or that I can learn from.

I love that.

You are getting a glimpse of what is to come.  Little bits of it you can run with and learn from if you choose to.  Plus of course we are here if you do not understand your classwork :)

Melody  Mar 3, 2016
#7
+93866
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I left out our physics people off my Credit list.  Alan is a Physicist and his posts sometimes blows my mind but Nauseated and Dragonlance as well as one or more guests often answer physics questions.  Physics is fascintating.  If there was an infinite amount of time I would definitely put effort into understanding physics.  :))

I have probably left some other great high end answerers off my list.... sorry.

Melody  Mar 3, 2016
#4
+91135
+5

cos2x - cosx = 0       using an identity, we have

2cos^2x - 1 - cosx  = 0   rearrange

2cos^2x - cosx - 1  =   0     factor

(2cosx + 1) (cosx - 1)  = 0

Setting each factor to 0, we have

2cosx + 1  = 0     subtract 1 from each side

2cosx  = -1      divide by 2 on each side

cos x  = -1/2       and this happens at x = 120° ± n360°   and at x = 240°  ± n360°  where n is a positive integer

For the other factor, we have

cos x - 1   = 0   add 1 to both sides

cos x = 1       and this is true at x = 0° ± n360°    where n is a positive integer

Here's a graph : https://www.desmos.com/calculator/bza7s6b83u

CPhill  Mar 3, 2016
#6
+93866
+5

Yep, CPhill's answer id the same as mine :)

Melody  Mar 3, 2016
#8
+20151
+10

cos2x - cosx = 0 i need steps....

rearrange

$$\begin{array}{rcll} \cos{(2x)} - \cos{(x)} &=& 0 \\ \cos{(2x)} &=& \cos{(x)} \\\\ \boxed{~ \cos{(\varphi)} = \cos{(-\varphi)}~}\\\\ \end{array}$$

We have 4 variations: {nl} $$\begin{array}{lrclrcl} (1) & \cos{(2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (2) & \cos{(-2x)} &=& \cos{(x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& x \pm 2k_2\pi\\ (3) & \cos{(2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\ (4) & \cos{(-2x)} &=& \cos{(-x)} \qquad \Rightarrow &\qquad -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\\\\ (1) & 2x \pm 2k_1\pi &=& x \pm 2k_2\pi\qquad \Rightarrow &\qquad x &=& 0 \pm 2k\pi\\ (2) & -2x \pm 2k_1\pi &=& x \pm 2k_2\pi \qquad \Rightarrow &\qquad -3x &=& 0 \pm 2k\pi\\ (3) & 2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad 3x &=& 0 \pm 2k\pi\\ (4) & -2x \pm 2k_1\pi &=& -x \pm 2k_2\pi\qquad \Rightarrow &\qquad -x &=& 0 \pm 2k\pi\\\\ & \Rightarrow x &=& \pm 2k\pi \\ & \Rightarrow x &=& \pm \frac{2}{3}k\pi \\ \end{array}$$

We can put together  $$x = \pm \frac{2}{3}k\pi \qquad k \in N_0$$

heureka  Mar 3, 2016
edited by heureka  Mar 3, 2016
edited by heureka  Mar 3, 2016