Giải phương trình $\frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{{{\sin }^6}x + {{\cos }^6}x}}{{4{{\cos }^2}2x + {{\sin }^2}2x}}$.
A. $x = k2\pi ,x = \frac{\pi }{2} + k2\pi $, $k \in \mathbb{Z}$.
B. $x = \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{2} + k\pi $, $k \in \mathbb{Z}$.
D. $x = k\pi ,x = \frac{\pi }{2} + k2\pi $, $k \in \mathbb{Z}$.
A. $x = k2\pi ,x = \frac{\pi }{2} + k2\pi $, $k \in \mathbb{Z}$.
B. $x = \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{2} + k\pi $, $k \in \mathbb{Z}$.
D. $x = k\pi ,x = \frac{\pi }{2} + k2\pi $, $k \in \mathbb{Z}$.
Chọn B.
Điều kiện: $4{\cos ^2}2x + {\sin ^2}2x \ne 0 \Leftrightarrow 4{\cos ^2}2x + 1 - {\cos ^2}2x \ne 0 \Leftrightarrow 3{\cos ^2}2x + 1 \ne 0 \Leftrightarrow \forall x \in \mathbb{R}$
$PT \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)}}{{4\left( {1 - {{\sin }^2}2x} \right) + {{\sin }^2}2x}}$
$ \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 3{{\sin }^2}x{{\cos }^2}x}}{{4 - 3{{\sin }^2}2x}}$
$ \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{1 - \frac{3}{4}{{\sin }^2}2x}}{{4 - 3{{\sin }^2}2x}} \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{4 - 3{{\sin }^2}2x}}{{4\left( {4 - 3{{\sin }^2}2x} \right)}}$
$ \Leftrightarrow {\sin ^{10}}x + {\cos ^{10}}x = 1 \Leftrightarrow {\sin ^{10}}x + {\cos ^{10}}x = {\sin ^2}x + {\cos ^2}x$
$ \Leftrightarrow {\sin ^2}x\left( {1 - {{\sin }^8}x} \right) + {\cos ^2}x\left( {1 - {{\cos }^8}x} \right) = 0\quad (*)$
Vì $\left\{ \begin{array}{l}{\sin ^2}x\left( {1 - {{\sin }^8}x} \right) \ge 0\forall x \in \mathbb{R}\\{\cos ^2}x\left( {1 - {{\cos }^8}x} \right) \ge 0\forall x \in \mathbb{R}\end{array} \right.$ nên $(*) \Leftrightarrow \left\{ \begin{array}{l}{\sin ^2}x\left( {1 - {{\sin }^8}x} \right) = 0\\{\cos ^2}x\left( {1 - {{\cos }^8}x} \right) = 0\end{array} \right.$$ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\sin x = 0\\\sin x = \pm 1\end{array} \right.\\\left[ \begin{array}{l}\cos x = 0\\\cos x = \pm 1\end{array} \right.\end{array} \right. \Leftrightarrow x = \frac{{k\pi }}{2}$
Điều kiện: $4{\cos ^2}2x + {\sin ^2}2x \ne 0 \Leftrightarrow 4{\cos ^2}2x + 1 - {\cos ^2}2x \ne 0 \Leftrightarrow 3{\cos ^2}2x + 1 \ne 0 \Leftrightarrow \forall x \in \mathbb{R}$
$PT \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)}}{{4\left( {1 - {{\sin }^2}2x} \right) + {{\sin }^2}2x}}$
$ \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 3{{\sin }^2}x{{\cos }^2}x}}{{4 - 3{{\sin }^2}2x}}$
$ \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{1 - \frac{3}{4}{{\sin }^2}2x}}{{4 - 3{{\sin }^2}2x}} \Leftrightarrow \frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{4 - 3{{\sin }^2}2x}}{{4\left( {4 - 3{{\sin }^2}2x} \right)}}$
$ \Leftrightarrow {\sin ^{10}}x + {\cos ^{10}}x = 1 \Leftrightarrow {\sin ^{10}}x + {\cos ^{10}}x = {\sin ^2}x + {\cos ^2}x$
$ \Leftrightarrow {\sin ^2}x\left( {1 - {{\sin }^8}x} \right) + {\cos ^2}x\left( {1 - {{\cos }^8}x} \right) = 0\quad (*)$
Vì $\left\{ \begin{array}{l}{\sin ^2}x\left( {1 - {{\sin }^8}x} \right) \ge 0\forall x \in \mathbb{R}\\{\cos ^2}x\left( {1 - {{\cos }^8}x} \right) \ge 0\forall x \in \mathbb{R}\end{array} \right.$ nên $(*) \Leftrightarrow \left\{ \begin{array}{l}{\sin ^2}x\left( {1 - {{\sin }^8}x} \right) = 0\\{\cos ^2}x\left( {1 - {{\cos }^8}x} \right) = 0\end{array} \right.$$ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\sin x = 0\\\sin x = \pm 1\end{array} \right.\\\left[ \begin{array}{l}\cos x = 0\\\cos x = \pm 1\end{array} \right.\end{array} \right. \Leftrightarrow x = \frac{{k\pi }}{2}$