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

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

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censoratojk
$$\displaystyle{\sin{{5}}}{x}={\cos{{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}}}={\cos{{3}}}{x}$$
$$\displaystyle{3}{x}={\frac{{\pi}}{{{2}}}}−{5}{x}+{2}{k}\pi$$
$$\displaystyle{x}={\frac{{\pi}}{{{16}}}}+{\frac{{{k}\pi}}{{{4}}}}={\frac{{\pi}}{{{16}}}}{\left({1}+{4}{k}\right)}$$
or
$$\displaystyle{3}{x}=−{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}+{2}{k}\pi$$
$$\displaystyle{x}={\frac{{\pi}}{{{4}}}}−{k}\pi$$
$$\displaystyle{x}={\frac{{\pi}}{{{4}}}}+{k}\pi={\frac{{\pi}}{{{4}}}}{\left({1}+{4}{k}\right)}$$
where $$\displaystyle{k}\in{Z}$$
writing $$\displaystyle-{k}\pi\ \text{ or }\ {k}\pi$$ does not change the solution set. Because −k is the opposite of k in integers.
###### Not exactly what you’re looking for?
Mary Nicholson
If you write m in place of n, you reached at $$\displaystyle{\frac{{\pi{\left({1}−{4}{m}\right)}}}{{{4}}}}$$
We
$$\displaystyle{\frac{{\pi{\left({1}−{4}{m}\right)}}}{{{4}}}}={\frac{{\pi{\left({1}+{4}{n}\right)}}}{{{4}}}}\Leftrightarrow{m}=-{n}$$
In our case m is any integer $$\displaystyle\Leftrightarrow{n}=−{m}$$ 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
$$\frac{\pi}{2}(1−4m)=\frac{\pi}{2}(4n+1),$$
so all that's really happened is tha tyou've listed the solutions in a different order.