Lời giải:

+) Ta có: \({\sin ^2}x = \frac{{1 - \cos 2x}}{2};{\cos ^2}x = \frac{{1 + \cos 2x}}{2}\)

\(P = {\sin ^{10}}x + {\cos ^{10}}x\)

\( = {\left( {{{\sin }^2}x} \right)^5} + {\left( {{{\cos }^2}x} \right)^5}\)

\( = \frac{{{{\left( {1 - \cos 2x} \right)}^5} + {{\left( {1 + \cos 2x} \right)}^5}}}{{{2^5}}}\)

\( = \frac{{\left( {1 - 5\cos 2x} \right) + 10{{\cos }^2}2x - 10{{\cos }^3}2x + 5{{\cos }^4}2x - {{\cos }^5}2x + \left( {1 + 5\cos 2x} \right) + 10{{\cos }^2}2x + 10{{\cos }^3}2x + 5{{\cos }^4}2x + {{\cos }^5}2x}}{{32}}\)

\( = \frac{{2 + 20{{\cos }^2}2x + 10{{\cos }^4}2x}}{{32}}\)

\( = \frac{5}{{16}}{\cos ^4}2x + \frac{5}{8}{\cos ^2}2x + \frac{1}{{16}}\)

+) \(P = \frac{1}{{16}} \Leftrightarrow \frac{5}{{16}}{\cos ^4}2x + \frac{5}{8}{\cos ^2}2x = 0\)

\( \Leftrightarrow \frac{5}{{16}}{\cos ^2}2x\left( {{{\cos }^2}2x + 2} \right) = 0\)

⇔ cos2x = 0  (do cos²2x + 2 > 0)

⇔ \(2x = \frac{\pi }{2} + k\pi \left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2}\left( {k \in \mathbb{Z}} \right)\).