1. The problem is to graph the function $y=3\cos(2\theta+6\pi)$.\n\n2. The general form of a cosine function is $y=A\cos(B\theta+C)$ where:\n- $A$ is the amplitude (height of the wave),\n- $B$ affects the period (how fast it oscillates),\n- $C$ is the phase shift (horizontal shift).\n\n3. For this function, $A=3$, $B=2$, and $C=6\pi$.\n\n4. The amplitude is $|3|=3$, so the graph oscillates between $-3$ and $3$.\n\n5. The period is given by $\frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi$. This means the function completes one full cycle every $\pi$ units of $\theta$.\n\n6. The phase shift is $-\frac{C}{B} = -\frac{6\pi}{2} = -3\pi$. This means the graph is shifted to the left by $3\pi$.\n\n7. The function can be rewritten using the phase shift as $y=3\cos\left(2\left(\theta+3\pi\right)\right)$.\n\n8. The graph is a cosine wave with amplitude 3, period $\pi$, shifted left by $3\pi$.\n\nFinal answer: The function to graph is $y=3\cos(2\theta+6\pi)$ with amplitude 3, period $\pi$, and phase shift $-3\pi$.