Pre-Calculus

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  $x = \pm \frac{2}{3}k\pi \qquad k \in N_0$

x would 0. 

= cos(2*0) - cos(0) 

= 1 - 1

= 0

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 ://

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