a,
pt \Leftrightarrow $cos\dfrac{4x}{3} = \dfrac{1 + cos2x}{2}$
\Leftrightarrow $2cos\dfrac{4x}{3} = 1 + cos3\dfrac{2}{3}x$
đặt $\dfrac{2}{3}x = t$
\Rightarrow $2cos2t = 1 + cos3t$
\Leftrightarrow $4cos^3t - 4cos^2t - 3cost + 3 = 0$
b.
Đặt $\left\{ \begin{array}{ll} \dfrac{3pi}{10} - \dfrac{x}{2} = a \\
\dfrac{3x}{2} + \dfrac{pi}{10} = b \\
\end{array} \right.$
\Rightarrow $\left\{ \begin{array}{ll} \dfrac{9pi}{10} - \dfrac{3x}{2} = 3a \\
\dfrac{3x}{2} + \dfrac{pi}{10} = b \\
\end{array} \right.$
\Rightarrow $pi = 3a + b$ \Rightarrow $ b = pi - 3a$
thay vào pt ta được
$sin(pi - 3a) = 3sina$
\Leftrightarrow $sin3a = 3sina$
\Leftrightarrow $3sina - 4sin^3a = 3sina$