Tìm $a$ để phương trình  $\sin2(x-\pi)-\sin(3x-\pi)=a\sin x$ có ít nhất một nghiệm $x \ne k\pi( {k \in Z} )$
Phương trình đã cho tương đương với
     $\sin 2{\rm{x +  sin3x}} = {\rm{a}}{\mathop{\rm s}\nolimits} {\rm{inx}}$
$ \Leftrightarrow {\mathop{\rm s}\nolimits} {\rm{inx}}\left( {4c{\rm{os}}2{\rm{x}} + 2c{\rm{osx}} - 1} \right) = {\rm{a}}{\mathop{\rm s}\nolimits} {\rm{inx}}$                           $(1)$
Do $x \ne k\pi  \Rightarrow (1) \Leftrightarrow 4c{\rm{o}}{{\rm{s}}^2}x + 2c{\rm{osx}} - 1 = a$            $(2)$
Đặt $t = c{\rm{osx}}$, $\left| t \right| < 1$ khi đó $(2)$ trở thành
        $f\left( t \right) = 4{t^2} + 2t - 1 = a,\left| t \right| < 1$
Hoành độ đỉnh của parabol $f\left( t \right)$ là ${t_0} = - \frac{2}{8} = - \frac{1}{4}$
Suy ra $\mathop {\min }\limits_{\left| t \right| \le 1} f\left( t \right) = f\left( {{t_0}} \right) = - \frac{5}{4},\mathop {m{\rm{ax}}}\limits_{\left| t \right| \le 1} f\left( t \right) = f\left( 1 \right) = 5$
Vậy để $f\left( t \right)$ có nghiệm $t$ : $\left| t \right| < 1$ cần có
$\mathop {\min }\limits_{\left| t \right| \le 1} f\left( t \right) \le a < \mathop {m{\rm{ax}}}\limits_{\left| t \right| \le 1} f\left( t \right) \Rightarrow  - \frac{5}{4} \le a < 5$
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