sin5x + 2cos²x = 1

\( \Leftrightarrow \) sin5x = 1 – 2cos²x

\( \Leftrightarrow \) sin5x = −cos2x = cos(−2x) = \(\sin \left( {\frac{\pi }{2} - \left( { - 2x} \right)} \right)\)

\( \Leftrightarrow \) sin5x = cos(−2x) = \(\sin \left( {\frac{\pi }{2} - \left( { - 2x} \right)} \right)\)

\( \Leftrightarrow \) sin5x = \(\sin \left( {\frac{\pi }{2} - \left( { - 2x} \right)} \right)\)

\( \Leftrightarrow \) sin5x \( = \sin \left( {2x + \frac{\pi }{2}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}5x = 2x + \frac{\pi }{2} + k2\pi \\5x = - 2x - \frac{\pi }{2} + k2\pi \end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + 2k\frac{\pi }{3}\\x = - \frac{\pi }{{14}} + k\frac{{2\pi }}{7}\end{array} \right.\) (k ∈ ℤ).