tìm x : sinx+2cosx/2=0

Đáp án:

\[x = \pi  + k2\pi \,\,\,\,\left( {k \in Z} \right)\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\sin x + 2\cos \dfrac{x}{2} = 0\\
 \Leftrightarrow 2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2} + 2.\cos \dfrac{x}{2} = 0\\
 \Leftrightarrow 2\cos \dfrac{x}{2}.\left( {\sin \dfrac{x}{2} + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{2} = 0\\
\sin \dfrac{x}{2} + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{2} = 0\\
\sin \dfrac{x}{2} =  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} = \dfrac{\pi }{2} + k\pi \\
\dfrac{x}{2} =  - \dfrac{\pi }{2} + k2\pi 
\end{array} \right. \Leftrightarrow \dfrac{x}{2} = \dfrac{\pi }{2} + k\pi  \Leftrightarrow x = \pi  + k2\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)