Bài tập 35 trang 42 SGK Toán 11 NC

a)

\(\begin{array}{*{20}{l}}
{{{\sin }^2}4x + {{\sin }^2}3x = {{\sin }^2}2x + {{\sin }^2}x}\\
\begin{array}{l}
 \Leftrightarrow \frac{1}{2}\left( {1 - \cos 8x} \right) + \frac{1}{2}\left( {1 - \cos 6x} \right)\\
 = \frac{1}{2}\left( {1 - \cos 4x} \right) + \frac{1}{2}\left( {1 - \cos 2x} \right)
\end{array}\\
{ \Leftrightarrow \cos 8x + \cos 6x = \cos 4x + \cos 2x}\\
{ \Leftrightarrow \cos 7x\cos x = \cos 3x\cos x}\\
{ \Leftrightarrow \cos x\left( {\cos 7x - \cos 3x} \right) = 0}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\cos x = 0}\\
{\cos 7x = \cos 3x}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = \frac{\pi }{2} + k\pi }\\
{x = k\frac{\pi }{2}}\\
{x = k\frac{\pi }{5}}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = k\frac{\pi }{2}}\\
{x = k\frac{\pi }{5}}
\end{array}} \right.\left( {k \in Z} \right)
\end{array}
\end{array}\)

b)

\(\begin{array}{l}
{\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x + {\cos ^2}4x = 2\\
 \Leftrightarrow \frac{{1 + \cos 2x}}{2} + \frac{{1 + \cos 4x}}{2} + \frac{{1 + \cos 6x}}{2} + \frac{{1 + \cos 8x}}{2} = 2\\
 \Leftrightarrow \left( {\cos 2x + \cos 4x} \right) + \left( {\cos 6x + \cos 8x} \right) = 0\\
 \Leftrightarrow 2\cos 3x\cos x + 2\cos 7x\cos x = 0\\
 \Leftrightarrow \cos x\left( {\cos 3x + \cos 7x} \right) = 0\\
 \Leftrightarrow 2\cos x\cos 5x\cos 2x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos 2x = 0\\
\cos 5x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k\pi \\
x = \frac{\pi }{4} + k\frac{\pi }{2}\\
x = \frac{\pi }{{10}} + k\frac{\pi }{5}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

-- Mod Toán 11 HỌC247