elvishwitchxyp

2022-01-02

Find, in radians the general solution of $\mathrm{cos}3x=\mathrm{sin}5x$
I have said, $\mathrm{sin}5x=\mathrm{cos}\left(\frac{\pi }{2}-5x\right)$
so
$\mathrm{cos}3x=\mathrm{sin}5x⇒3x=2n\pi ±\left(\frac{\pi }{2}-5x\right)$
When I add I get the answer $x=\frac{\pi }{16}\left(4n+1\right)$, which the book says is correct.
But when I subtract I get a different answer to the book. My working is as follows:
$3x=2n\pi -\frac{\pi }{2}+5x$
$2x=\frac{\pi }{2}-2n\pi$
$x=\frac{\pi }{4}-n\pi =\frac{\pi }{4}\left(1-4n\right)$

censoratojk

$\mathrm{sin}5x=\mathrm{cos}\left(\frac{\pi }{2}-5x\right)=\mathrm{cos}3x$
$3x=\frac{\pi }{2}-5x+2k\pi$
$x=\frac{\pi }{16}+\frac{k\pi }{4}=\frac{\pi }{16}\left(1+4k\right)$
or
$3x=-\left(\frac{\pi }{2}-5x\right)+2k\pi$
$x=\frac{\pi }{4}-k\pi$
$x=\frac{\pi }{4}+k\pi =\frac{\pi }{4}\left(1+4k\right)$
where $k\in Z$
writing does not change the solution set. Because −k is the opposite of k in integers.

Mary Nicholson

If you write m in place of n, you reached at $\frac{\pi \left(1-4m\right)}{4}$
We
$\frac{\pi \left(1-4m\right)}{4}=\frac{\pi \left(1+4n\right)}{4}⇔m=-n$
In our case m is any integer $⇔n=-m$ also belong to the same infinite set of integers
In their case n is so.

Vasquez

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

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