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cos(14x)+cos(12x)

 Mar 20, 2020
edited by rubikx2910  Apr 15, 2020
edited by Melody  Jan 10, 2022

Best Answer 

 #6
avatar+118608 
+2

You can use the identity 

 

\(\boxed{cos(a)+cos(b)=2\;cos\frac{a+b}{2}\;cos\frac{a-b}{2}}\\~\\\)

\(y = \cos(12x) + \cos(14x)\\ \text{roots are given by}\\ \cos(12x) + \cos(14x)=0\\ 2\cos\frac{14x+12x}{2}\;\cos \frac{14x-12x}{2}=0\\ 2\cos(13x)\;\cos (x)=0\\ \cos(13x)=0\qquad \qquad  or \qquad cos(x)=0\\ 13x=\pm \frac{\pi}{2}+2\pi n \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\ x=\pm \frac{\pi}{26}+\frac{2\pi n}{13} \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\ x=\pm \frac{\pi}{26}+\frac{4\pi n}{26} \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\ x=\frac{\pi}{26}(4n\pm1) \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\ \)

 

n 0112233
x \(\frac{\pi}{26}\)\(\frac{3\pi}{26}\)\(\frac{5\pi}{26}\)\(\frac{7\pi}{26}\)\(\frac{9\pi}{26}\)\(\frac{11\pi}{26}\)etc

 

 

Coding:

y = \cos(12x) + \cos(14x)\\
\text{roots are given by}\\
\cos(12x) + \cos(14x)=0\\
2\cos\frac{14x+12x}{2}\;\cos \frac{14x-12x}{2}=0\\
2\cos(13x)\;\cos (x)=0\\
\cos(13x)=0\qquad \qquad  or \qquad cos(x)=0\\
13x=\pm \frac{\pi}{2}+2\pi n \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\
x=\pm \frac{\pi}{26}+\frac{2\pi n}{13} \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\
x=\pm \frac{\pi}{26}+\frac{4\pi n}{26} \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\
x=\frac{\pi}{26}(4n\pm1) \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\

 Mar 22, 2020
 #6
avatar+118608 
+2
Best Answer

You can use the identity 

 

\(\boxed{cos(a)+cos(b)=2\;cos\frac{a+b}{2}\;cos\frac{a-b}{2}}\\~\\\)

\(y = \cos(12x) + \cos(14x)\\ \text{roots are given by}\\ \cos(12x) + \cos(14x)=0\\ 2\cos\frac{14x+12x}{2}\;\cos \frac{14x-12x}{2}=0\\ 2\cos(13x)\;\cos (x)=0\\ \cos(13x)=0\qquad \qquad  or \qquad cos(x)=0\\ 13x=\pm \frac{\pi}{2}+2\pi n \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\ x=\pm \frac{\pi}{26}+\frac{2\pi n}{13} \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\ x=\pm \frac{\pi}{26}+\frac{4\pi n}{26} \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\ x=\frac{\pi}{26}(4n\pm1) \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\ \)

 

n 0112233
x \(\frac{\pi}{26}\)\(\frac{3\pi}{26}\)\(\frac{5\pi}{26}\)\(\frac{7\pi}{26}\)\(\frac{9\pi}{26}\)\(\frac{11\pi}{26}\)etc

 

 

Coding:

y = \cos(12x) + \cos(14x)\\
\text{roots are given by}\\
\cos(12x) + \cos(14x)=0\\
2\cos\frac{14x+12x}{2}\;\cos \frac{14x-12x}{2}=0\\
2\cos(13x)\;\cos (x)=0\\
\cos(13x)=0\qquad \qquad  or \qquad cos(x)=0\\
13x=\pm \frac{\pi}{2}+2\pi n \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\
x=\pm \frac{\pi}{26}+\frac{2\pi n}{13} \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\
x=\pm \frac{\pi}{26}+\frac{4\pi n}{26} \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\
x=\frac{\pi}{26}(4n\pm1) \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\

Melody Mar 22, 2020

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