Bài tập 1.14 trang 23 SBT Toán 11

a) \(\sin 3x =  - \frac{{\sqrt 3 }}{2}\)

\( \Leftrightarrow \sin 3x = \sin ( - \frac{\pi }{3})\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{3x =  - \frac{\pi }{3} + k2\pi ,k \in Z}\\
{3x = \pi  - ( - \frac{\pi }{3}) + k2\pi ,k \in Z}
\end{array}} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}
x =  - \frac{\pi }{9} + k\frac{{2\pi }}{3},k \in Z\\
x = \frac{{4\pi }}{9} + k\frac{{2\pi }}{3},k \in Z
\end{array} \right.\)

b) \(\sin (2x - {15^o}) = \frac{{\sqrt 2 }}{2}\)

\( \Leftrightarrow \sin (2x - {15^o}) = \sin ({45^o})\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{2x - {{15}^o} = {{45}^o} + k{{360}^o},k \in Z}\\
{2x - {{15}^o} = {{135}^o} + k{{360}^o},k \in Z}
\end{array}} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}
x = {30^o} + k{180^o},k \in Z\\
x = {75^o} + k{180^o},k \in Z
\end{array} \right.\)

c) \(\sin (\frac{x}{2} + {10^o}) =  - \frac{1}{2}\)

\( \Leftrightarrow \sin (\frac{x}{2} + {10^o}) = \sin ( - {30^o})\)

\(\frac{x}{2} + {10^o} =  - {30^o} + k{360^o},k \in Z\frac{x}{2} + {10^o} = {210^o} + k{360^o},k \in Z\)

\( \Leftrightarrow \left[ \begin{array}{l}
x =  - {80^o} + k{720^o},k \in Z\\
x = {400^o} + k{720^o},k \in Z
\end{array} \right.\)

d) \(\sin 4x = \frac{2}{3}\)

\( \Leftrightarrow \sin 4x = \sin (\arcsin \frac{2}{3})\)

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

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

-- Mod Toán 11 HỌC247