Bài tập 34 trang 42 SGK Toán 11 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos x\cos 5x = \cos 2x\cos 4x\\
 \Leftrightarrow \frac{1}{2}\left( {\cos 6x + \cos 4x} \right) = \frac{1}{2}\left( {\cos 6x + \cos 2x} \right)
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \cos 4x = \cos 2x \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{4x = 2x + k2\pi }\\
{4x =  - 2x + k2\pi }
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = k\pi }\\
{x = k\frac{\pi }{3}}
\end{array}} \right. \Leftrightarrow x = k\frac{\pi }{3}\left( {k \in Z} \right)
\end{array}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos 5x\sin 4x = \cos 3x\sin 2x\\
 \Leftrightarrow \frac{1}{2}\left( {\sin 9x - \sin x} \right) = \frac{1}{2}\left( {\sin 5x - \sin x} \right)
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \sin 9x = \sin 5x \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{9x = 5x + k2\pi }\\
{9x = \pi  - 5x + k2\pi }
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = k\frac{\pi }{2}}\\
{x = \frac{\pi }{{14}} + k\frac{\pi }{7}}
\end{array}} \right.\left( {k \in Z} \right)
\end{array}
\end{array}\)

c)

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

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin x + \sin 2x = \cos x + \cos 2x\\
 \Leftrightarrow 2\sin \frac{{3x}}{2}\cos \frac{x}{2} = 2\cos \frac{{3x}}{2}\cos \frac{x}{2}
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \cos \frac{x}{2}\left( {\sin \frac{{3x}}{2} - \cos \frac{{3x}}{2}} \right) = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\cos \frac{x}{2} = 0}\\
{\sin \frac{{3x}}{2} = \cos \frac{{3x}}{2}}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\
{\tan \frac{{3x}}{2} = 1}
\end{array}} \right.
\end{array}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = \pi  + k2\pi }\\
{x = \frac{\pi }{6} + k\frac{{2\pi }}{3}}
\end{array}} \right.\left( {k \in Z} \right)}
\end{array}\)

-- Mod Toán 11 HỌC247