What is the period of y = 2 cos(3x - 2 pi)?

Step-by-step solution

1. 1

   Identify the three parameters. Compare with the general form $y = A\cos(bx - c)$: the amplitude is $A = 2$, the coefficient of $x$ is $b = 3$, and the phase term is $c = 2\pi$. Only one of these three controls the period.

2. 2

   Use the period formula. Cosine completes one cycle when its argument advances by $2\pi$. Here the argument is $3x - 2\pi$, which advances by $2\pi$ when $x$ advances by $2\pi/3$:

   $3(x + p) - 2\pi = (3x - 2\pi) + 2\pi \;\Longrightarrow\; 3p = 2\pi \;\Longrightarrow\; p = \frac{2\pi}{3}$

3. 3

   See why the amplitude does not matter. The factor $2$ scales the output vertically. Since $2\cos\theta$ repeats exactly when $\cos\theta$ repeats, the horizontal cycle length is unaffected — a taller wave is not a longer wave.

4. 4

   See why the phase shift does not matter either. Subtracting $2\pi$ inside slides the graph right by $2\pi/3$ units, but sliding a periodic graph never changes how often it repeats. In fact $2\pi$ is a full cosine period, so here $\cos(3x - 2\pi) = \cos 3x$ and the graph is not even shifted visibly.

5. 5

   State and check the answer. The period is $\dfrac{2\pi}{3}$. Verifying directly: $y(0) = 2\cos(-2\pi) = 2$, and $y\bigl(\tfrac{2\pi}{3}\bigr) = 2\cos(2\pi - 2\pi) = 2\cos 0 = 2$ ✓, while the midpoint $y\bigl(\tfrac{\pi}{3}\bigr) = 2\cos(\pi - 2\pi) = -2$ shows a genuine full swing in between. Distractors like $3\pi$ or $6\pi$ come from multiplying by $b$ instead of dividing.