Find, in radians the general solution of cos⁡3x=sin⁡5x I have said, sin⁡5x=cos⁡(π2−5x) so cos⁡3x=sin⁡5x⇒3x=2nπ±(π2−5x) When...

elvishwitchxyp

elvishwitchxyp

Answered

2022-01-02

Find, in radians the general solution of cos3x=sin5x
I have said, sin5x=cos(π25x)
so
cos3x=sin5x3x=2nπ±(π25x)
When I add (π25x)  to  2nπ I get the answer x=π16(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=2nππ2+5x
2x=π22nπ
x=π4nπ=π4(14n)

Answer & Explanation

censoratojk

censoratojk

Expert

2022-01-03Added 46 answers

sin5x=cos(π25x)=cos3x
3x=π25x+2kπ
x=π16+kπ4=π16(1+4k)
or
3x=(π25x)+2kπ
x=π4kπ
x=π4+kπ=π4(1+4k)
where kZ
writing kπ  or  kπ does not change the solution set. Because −k is the opposite of k in integers.
Mary Nicholson

Mary Nicholson

Expert

2022-01-04Added 38 answers

If you write m in place of n, you reached at π(14m)4
We
π(14m)4=π(1+4n)4m=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

Vasquez

Expert

2022-01-09Added 457 answers

No, the two are equivalent. In particular, if m = −n, then
π2(14m)=π2(4n+1),
so all that's really happened is tha tyou've listed the solutions in a different order.

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