What is the period of y=2cos(3x-2π)?

Mathematics

1 answer:

Strike441 [17]3 years ago

3 0

$y=2\cos(3x-2\pi)\\\\\text{The period of a cosine is}\ 2\pi.\\\\P-period\\\\2\cos(3(x+P)-2\pi)=2\cos(3x-2\pi)\qquad\text{divide both sides by 2}\\\\\cos(3x+3P-2\pi)=\cos(3x-2\pi)\iff3x+3P-2\pi=3x-2\pi+2\pi\\\\3x+3P-2\pi=3x\qquad\text{subtract 3x from both sides}\\\\3P-2\pi=0\qquad\text{add}\ 2\pi\ \text{to both sides}\\\\3P=2\pi\qquad\text{divide both sides by 3}\\\\\boxed{P=\dfrac{2\pi}{3}}\\\\Answer:\ \boxed{D.\ \dfrac{2\pi}{3}}$

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Answer:

Step-by-step explanation:

Given system is

$\left \{ {{x+y=0} \atop {x-y+2=0}} \right.$

If we graph, the solution would be the interception point between these two lines, because each linear equation represents a line. The lines are attached.

In the graph, you can observe that the solution is the point (-1,1), that is x = -1 and y = 1.

Now, if we want to solve this system by another method, we can just sum both equations and solve for <em>x</em>

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