Bài tập 14 trang 28 SGK Toán 11 NC

a)

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

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin \left( {\frac{{x + \pi }}{5}} \right) =  - \frac{1}{2}\\
 \Leftrightarrow \sin \left( {\frac{{x + \pi }}{5}} \right) = \sin \left( { - \frac{\pi }{6}} \right)
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\frac{{x + \pi }}{5} =  - \frac{\pi }{6} + k2\pi }\\
{\frac{{x + \pi }}{5} = \pi  + \frac{\pi }{6} + k2\pi }
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x =  - \frac{{11\pi }}{6} + k10\pi }\\
{x = \frac{{29\pi }}{6} + k10\pi }
\end{array}} \right.
\end{array}
\end{array}\)

c)

\(\begin{array}{l}
\cos \frac{x}{2} = \cos \sqrt 2  \Leftrightarrow \frac{x}{2} =  \pm \sqrt 2  + k2\pi \\
 \Leftrightarrow x =  \pm 2\sqrt 2  + k4\pi \left( {k \in Z} \right)
\end{array}\)

d) Vì \(0 < \frac{2}{5} < 1\) nên có số \(\alpha \) sao cho \(\cos \alpha  = \frac{2}{5}\).

Do đó:

\(\begin{array}{l}
(\cos \left( {x + \frac{\pi }{{18}}} \right) = \frac{2}{5} \Leftrightarrow \cos \left( {x + \frac{\pi }{{18}}} \right) = \cos \alpha \\
 \Leftrightarrow x =  \pm \alpha  - \frac{\pi }{{18}} + k2\pi ,k \in Z
\end{array}\)

-- Mod Toán 11 HỌC247