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?
$\endgroup$ 13 Answers
$\begingroup$$$\sin 5x = \cos (\frac{\pi}{2}-5x)= \cos 3x $$
$$3x=\frac{\pi}{2}-5x+2k\pi$$
$$x=\frac{\pi}{16}+\frac{k\pi}{4}=\frac{\pi}{16}(1+4k)$$
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.
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ $\begingroup$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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