Giải thích các bước giải:
b. 3cosx+5sin$\frac{x}{2}$ -4=0
<-> 3(1-2sin²$\frac{x}{2}$ )+5sin$\frac{x}{2}$ -4=0
<-> -6sin²$\frac{x}{2}$+5sin$\frac{x}{2}$-1=0
<-> \(\left[ \begin{array}{l}
\sin \frac{x}{2} = \frac{1}{2}\\
\sin \frac{x}{2} = \frac{1}{3}
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
\frac{x}{2} = \frac{\pi }{6} + k2\pi \\
\frac{x}{2} = \frac{{5\pi }}{6} + k2\pi \\
\frac{x}{2} = \arcsin \frac{1}{3} + k2\pi \\
\frac{x}{2} = \pi - \arcsin \frac{1}{3} + k2\pi
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{3} + k2\pi \\
\frac{x}{2} = \frac{{5\pi }}{3} + k2\pi \\
x = 2\arcsin \frac{1}{3} + k2\pi \\
x = 2\pi - 2\arcsin \frac{1}{3} + k2\pi
\end{array} \right.(k \in Z)\)
