$\begin{array}{l}
a)\,\,\,{\sin ^2}2x + 3\sin 2x + 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = - 1\\
\sin 2x = - 2\,\,\,\left( {ktm} \right)
\end{array} \right. \Leftrightarrow 2x = - \frac{\pi }{2} + k2\pi \Leftrightarrow x = - \frac{\pi }{4} + k\pi \,\,\left( {k \in Z} \right)\\
x \in \left[ {0;\,\,10\pi } \right] \Rightarrow 0 \le - \frac{\pi }{4} + k\pi \le 10\pi \\
\Leftrightarrow \frac{\pi }{4} \le k\pi \le \frac{{41\pi }}{4} \Leftrightarrow 0,25 \le k \le 10,25\\
\Rightarrow k \in \left\{ {1;\,\,2;\,\,3;\,\,4;\,\,5;\,\,6;\,\,7;\,\,8;\,\,10} \right\}\\
\Rightarrow x \in \left\{ {\frac{\pi }{4};\,\frac{{3\pi }}{4};\,....\,;\,\,\frac{{19\pi }}{4}} \right\}\\
S = \frac{\pi }{4} + \frac{{3\pi }}{4} + \,....\, + \,\frac{{19\pi }}{4} = ...\\
b)\,\,\,2{\sin ^2}x - 5\sin x + 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 2\,\,\,\left( {ktm} \right)\\
\sin x = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + m2\pi
\end{array} \right.\,\,\,\left( {k,\,\,m \in Z} \right)\\
Nghiem\,\,am\,\,lon\,\,nhat \Rightarrow \left[ \begin{array}{l}
\frac{\pi }{6} + k2\pi < 0\\
\frac{{5\pi }}{6} + m2\pi < 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
k < - \frac{1}{{12}}\\
m < - \frac{5}{{12}}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
k = - 1\\
m = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - \frac{{11\pi }}{6}\\
x = - \frac{{7\pi }}{6}
\end{array} \right. \Rightarrow Nghiem\,\,am\,\,lon\,\,nhat\,\,la:\,\,\,x = - \frac{{7\pi }}{6}.
\end{array}$
Ý c em làm tiếp nhé!
