+26367 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 \(x = \pm \frac{2}{3}k\pi \qquad k \in N_0\)
+118608 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 ://
+118608 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 :)
+118608 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.
+128408 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
+26367 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 \(x = \pm \frac{2}{3}k\pi \qquad k \in N_0\)
