elvishwitchxyp

2022-01-02

Find, in radians the general solution of $\mathrm{cos}3x=\mathrm{sin}5x$

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

so

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

When I add$(\frac{\pi}{2}-5x)\text{}\text{to}\text{}2n\pi$ 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:

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

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

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

I have said,

so

When I add

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

censoratojk

Beginner2022-01-03Added 46 answers

or

where

writing

Mary Nicholson

Beginner2022-01-04Added 38 answers

If you write m in place of n, you reached at $\frac{\pi (1-4m)}{4}$

We

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

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.

We

In our case m is any integer

In their case n is so.

Vasquez

Expert2022-01-09Added 669 answers

No, the two are equivalent. In particular, if m = −n, then

so all that's really happened is tha tyou've listed the solutions in a different order.

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