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In one of my high school maths questions the example given to find the maximum displacement of a Simple Harmonic Motion where $ x=2+4\cos \left (2t + \frac{\pi}{3} \right ) $ and the motion lies in the interval: $-2 \leq x\leq 6$ is:

  1. let $x=2+4\cos \left (2t + \frac{\pi}{3} \right )= 6$
  2. then $\cos \left (2t + \frac{\pi}{3} \right )= 1$
  3. $2t + \frac{\pi}{3} = 2\pi$
  4. $t = \frac{5\pi}{6}$

I am getting confused on the transition from line 2 to line 3. I'm confused as to how getting rid of the cos turns 1 into $2\pi$. Could someone please explain this at a high school level?

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up vote 3 down vote accepted

Remember: $$\cos x=1\Longleftrightarrow x=2k\pi\,,\,k\,\,\text{an integer}$$so$$\cos\left(2t+\frac{\pi}{3}\right)=1\Longleftrightarrow 2t+\frac{\pi}{3}=2k\pi$$

Assuming, as surely is the case, that it must be $\,t\geq 0\,$ , we get that$$2k\pi>\frac{\pi}{3}\Longrightarrow k=1,2,3,...$$and we can choose $k=1\Longrightarrow 2t+\frac{\pi}{3}=2\pi$

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You normally would apply the $\arccos$ function to both sides - see here. This would yield $2t+\tfrac {\pi} 3=0$, however, and the question implies (or perhaps tacitly assumes) a positive value of $t$. Note that, observing the cosine function, we see that it equals $1$ whenever it's argument is $0,2\pi,4\pi,...$ in general whenever it's argument is $n\cdot 2\pi$, where $n$ is an integer.

There are actually infinitely many values of $t$ for which $\cos(2t+\tfrac{\pi} 3)=1$, but the one provided is the smallest positive value.

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