\(\displaystyle{\sin{{5}}}{x}={\cos{{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}}}={\cos{{3}}}{x}\)

\(\displaystyle{3}{x}={\frac{{\pi}}{{{2}}}}−{5}{x}+{2}{k}\pi\)

\(\displaystyle{x}={\frac{{\pi}}{{{16}}}}+{\frac{{{k}\pi}}{{{4}}}}={\frac{{\pi}}{{{16}}}}{\left({1}+{4}{k}\right)}\)

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.

\(\displaystyle{3}{x}={\frac{{\pi}}{{{2}}}}−{5}{x}+{2}{k}\pi\)

\(\displaystyle{x}={\frac{{\pi}}{{{16}}}}+{\frac{{{k}\pi}}{{{4}}}}={\frac{{\pi}}{{{16}}}}{\left({1}+{4}{k}\right)}\)

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.