Bài tập 1.58 trang 41 SBT Toán 11

Ta có: \(8{\sin ^6}x = {\sin ^2}2x\)

\(\begin{array}{l}
 \Leftrightarrow 8{\sin ^6}x = 4{\sin ^2}x{\cos ^2}x\\
 \Leftrightarrow 4{\sin ^2}x(2{\sin ^4}x + {\sin ^2}x - 1) = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\sin }^2}x = 0}\\
{2{{\sin }^4}x + {{\sin }^2}x - 1 = 0}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = k\pi ,k \in Z}\\
{{{\sin }^2}x = \frac{1}{2}}\\
{{{\sin }^2}x =  - 1 \le 0\,\,{\rm{(l)}}}
\end{array}} \right.
\end{array}\)

\(\begin{array}{l}
{\sin ^2}x = \frac{1}{2}\\
 \Leftrightarrow \frac{{1 - \cos 2x}}{2} = \frac{1}{2}\\
 \Leftrightarrow \cos 2x = 0\\
 \Leftrightarrow 2x = \frac{\pi }{2} + k\pi ,k \in Z\\
 \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2},k \in Z
\end{array}\)

Đáp án: D.

-- Mod Toán 11 HỌC247