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We have domain $\,[0, 2\pi]\,$ and the following functions are given: $$f(x)=\cos(2x) \text{ and } g(x)=\sin(x-\pi/3)$$Solve exactly: $\,f(x)=g(x)$

Why does one solve: (right way) $$\cos(2x)=\sin(x-\pi/3)\\\cos(2x)=\cos(\pi/2-(x-\pi/3))\\ \text{etc.}\ldots $$ and not (which gives the wrong answer $\rightarrow a - k\,\, \times\,\, 2\pi \cdots$): (wrong way) $$\cos(2x)=\sin(x-\pi/3)\sin(\pi/2-(2x))=\sin(x-\pi/3)\\ \text{etc.}\ldots$$

Right way fully worked out:

$$\cos(2x) = \sin(x -\pi/3)\\ \cos(2x) = \cos(\pi/2 - (x -\pi/3))\\ \cos(2x) = \cos(\pi/2 - x + \pi/3)\\ \cos(2x) = \cos(5\pi/6 - x)\\ 2x = (5\pi/6 - x) + k\,\, \times \,\, 2\pi\\ 3x = 5\pi / 6 + k \,\, \times \,\, 2\pi\\ x = 5\pi /18 + k \,\, \times \,\, 2\pi\\ or:\\ 2x = - (5\pi/6 - x) + k\,\, \times \,\, 2\pi\\ x = -5\pi / 6 + k \,\, \times \,\, 2\pi\\ $$

Wrong way fully worked out:

$$\cos(2x) = \sin(x-\pi/3)\\ sin(\pi/2 - x)=\sin(x-\pi/3)\\ \pi/2-2x=(x-\pi/3)+k\,\, \times \,\, 2\pi\\ -3x = (-\pi/3 - \pi/2) + k\,\, \times \,\, 2\pi\\ -3x = -5\pi/6+ k\,\, \times \,\, 2\pi\\ x = 5\pi/18 -k \,\, \times \,\, 2\pi\\ or:\\ \pi/2-2x=\pi-(x-\pi/3) + k \,\, \times \,\, 2\pi\\ \pi/2 - x = \pi - x + \pi/3 + k \,\, \times \,\, 2\pi\\ x= 4\pi/3-\pi/2 + k \,\, \times \,\, 2\pi\,\,\,\\\ \,\,\, =5\pi/6 - k \,\, \times \,\,2\pi $$

On closer inspection of the worked out examples above you can clearly see that final answers of the wrong way are.. wrong. This because the answers we get aren't within the set domain.

Help is highly appreciated.

-Bowser

Cro-Magnon
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  • I don't understand what do you mean with $a-k \times 2 \pi $ – Skills Mar 14 '16 at 11:10
  • When you solve the right way you get: $a + k ,,,\times,,, 2\pi$ which gives you all the solution you need. When you solve the wrong way you get $a - k ,,,\times,,, 2\pi$. You see? because of the minus sign the answers you get are wrong. – Cro-Magnon Mar 14 '16 at 11:16
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    We don't know what you mean by $a$, and we don't know what you mean by $k$. But if $k$ is supposed to run through the integers, then $a-2\pi k$ is exactly the same set of numbers as $a+2\pi k$, so both methods give you the same answer. – Gerry Myerson Mar 14 '16 at 11:31
  • I'll write it out. – Cro-Magnon Mar 14 '16 at 14:44
  • @gerryMyerson I've worked them out. – Cro-Magnon Mar 15 '16 at 12:38
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    You have "worked them out", but the symbol $k$ has flown into your work from the blue, without any indication of what it means. My previous comment stands. Also, from $\cos v=\cos w$ (or $\sin v=\sin w$), you cannot draw the conclusion that $v$ is $w$ plus an integer multiple of $2\pi$. E.g., $\sin(\pi/6)=\sin(5\pi/6)$, but there is no integer $k$ such that $\pi/6=5\pi/6+2k\pi$. I think you have to go back to the drawing board. – Gerry Myerson Mar 15 '16 at 21:57
  • $k$ is simply the next solution.. This is not something I'm making up. Maybe you've just been taught differently. I'll wait for someone else thanks anyway. – Cro-Magnon Mar 15 '16 at 22:14
  • I have been taught that when you introduce a symbol, you tell people what that symbol stands for. I'm glad someone else came along to give you the answer on a silver platter, but, really, you would have done much better to think seriously about and engage with the comments I made. – Gerry Myerson Mar 16 '16 at 06:31
  • Thanks for the rebuke. I'll try harder. – Cro-Magnon Mar 16 '16 at 09:09

1 Answers1

1

You can choose to transform the sine into cosine or conversely.

First method

\begin{gather} \cos2x=\cos\left(\frac{\pi}{2}-\left(x-\frac{\pi}{3}\right)\right) \\[6px] \cos2x=\cos\left(\frac{5\pi}{6}-x\right) \\[6px] 2x=\frac{5\pi}{6}-x+2k\pi \qquad\text{or}\qquad 2x=-\frac{5\pi}{6}+x+2k\pi \\[6px] x=\frac{5\pi}{18}+k\frac{2\pi}{3} \qquad\text{or}\qquad x=-\frac{5\pi}{6}+2k\pi \end{gather}

Second method

\begin{gather} \sin\left(\frac{\pi}{2}-2x\right)=\sin\left(x-\frac{\pi}{3}\right) \\[6px] \frac{\pi}{2}-2x=x-\frac{\pi}{3}+2k\pi \qquad\text{or}\qquad \frac{\pi}{2}-2x=\pi-x+\frac{\pi}{3}+2k\pi \\[6px] x=\frac{5\pi}{18}-k\frac{2\pi}{3} \qquad\text{or}\qquad x=-\frac{5\pi}{6}-2k\pi \end{gather}

What's happening?

Nothing strange. In the solutions above, $k$ denotes an arbitrary integer: for any choice of the integer $k$, you get a solution. So writing $-k$ or $+k$ is irrelevant: the solution for $k=1$ in the first set (on the left side) appears in the solutions of the second set (left side) for $k=-1$.

The two forms give the same solutions.

If you want to find the solutions in the interval $[0,2\pi)$ you can do as follows, for the first method: $$ 0\le\frac{5\pi}{18}+k\frac{2\pi}{3}<2\pi \iff 0\le5+12k<36 \iff -5\le12k<31 $$ that gives $k=0,1,2$. Also $$ 0\le-\frac{5\pi}{6}+2k\pi<2\pi \iff 0\le-5+12k<12 \iff 5\le12k<17 $$ that gives $k=1$. So you have four solutions.

The same procedure in the second method would give $k=0,-1,-2$ in one case and $k=-1$ in the other.

egreg
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