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I am studying maths as a hobby. I have come across this problem:

Find a general solution for the equation cos 3x = sin 5x

I have said, $\sin 5x = \cos(\frac{\pi}{2} - 5x)$

so

$\cos 3x = \sin 5x \implies 3x = 2n\pi\pm(\frac{\pi}{2} - 5x)$

When I add $(\frac{\pi}{2} - 5x)$ to $2n\pi$ I get the answer $x = \frac{\pi}{16}(4n +1)$, which the book says is correct.

But when I subtract I get a different answer to the book. My working is as follows:

$3x = 2n\pi - \frac{\pi}{2} + 5x$

$2x = \frac{\pi}{2} - 2n\pi$

$x = \frac{\pi}{4} - n\pi = \frac{\pi}{4}(1 - 4n)$

but my text book says the answer is $\frac{\pi}{4}(4n + 1)$

Is the book wrong?

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$$\sin 5x = \cos (\frac{\pi}{2}-5x)= \cos 3x $$

$$3x=\frac{\pi}{2}-5x+2k\pi$$

$$x=\frac{\pi}{16}+\frac{k\pi}{4}=\frac{\pi}{16}(1+4k)$$

or

$$3x=-(\frac{\pi}{2}-5x)+2k\pi$$

$$x=\frac{\pi}{4}-k\pi$$

$$x=\frac{\pi}{4}+k\pi =\frac{\pi}{4}(1+4k)$$

where $k\in Z$

writing $-k\pi$ or $k\pi$ does not change the solution set. Because $-k$ is the opposite of $k$ in integers.

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  • $\begingroup$ I'm not sure about the last sentence. I would have thought that because -k is the opposite of k then it must change the solution $\endgroup$ – Steblo Dec 3 '20 at 18:12
  • $\begingroup$ I mean it will give the same solution set by writing $-k$ for $k$ $\endgroup$ – Lion Heart Dec 3 '20 at 18:15
  • $\begingroup$ Is that because cos -x is the same as cosx? $\endgroup$ – Steblo Dec 3 '20 at 18:25
  • $\begingroup$ @Steblo : take $x= \frac{\pi}{4}(1-4k)$ for $k=1$, $x=-\frac {3pi}{4}$. Now take take $x= \frac{\pi}{4}(1+4k)$ for $k=-1$, $x=-\frac {3pi}{4}$. $\endgroup$ – Lion Heart Dec 3 '20 at 18:32
  • $\begingroup$ @Steblo : $\cos(-x)= \cos x$. Because Cosine function is an even function and its graph is symmetric to the y-axis or it can be say different reasons. $\endgroup$ – Lion Heart Dec 3 '20 at 18:43
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If you write $m$ in place of $n,$ you reached at $\dfrac{\pi(1-4m)}4$

We $$\dfrac{\pi(1-4m)}4=\dfrac{\pi(1+4n)}4\iff m=-n$$

In our case $m$ is any integer $\iff n=-m$ also belong to the same infinite set of integers

In their case $n$ is so.

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No, the two are equivalent. In particular, if $m$ = $-n$, then $$\dfrac{\pi}{2}(1 - 4m) = \dfrac{\pi}{2}(4n + 1),$$ so all that's really happened is tha tyou've listed the solutions in a different order.

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