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 I add (π2−5x) 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=2nπ−π2+5x 2x=π2−2nπ x=π4−nπ=π4(1−4n)

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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 I add (π2−5x)  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=2nπ−π2+5x  2x=π2−2nπ  x=π4−nπ=π4(1−4n)

censoratojk · Accepted Answer

sin⁡5x=cos⁡(π2−5x)=cos⁡3x   3x=π2−5x+2kπ   x=π16+kπ4=π16(1+4k)   or  3x=−(π2−5x)+2kπ   x=π4−kπ   x=π4+kπ=π4(1+4k)   where k∈Z  writing −kπ  or  kπ does not change the solution set. Because −k is the opposite of k in integers.

Mary Nicholson · Accepted Answer

If you write m in place of n, you reached at π(1−4m)4   We   π(1−4m)4=π(1+4n)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 · Accepted Answer

No, the two are equivalent. In particular, if m = −n, then π2(1−4m)=π2(4n+1), so all that&#039;s really happened is tha tyou&#039;ve listed the solutions in a different order.

Find, in radians the general solution of \cos 3x = \sin 5

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