a) Ta có:
$\quad \begin{cases}\sin^2\alpha + \cos^2\alpha = 1\\\sin\alpha +\cos\alpha = 1,4\end{cases}$
$\Leftrightarrow \begin{cases}\sin\alpha = 1,4 -\cos\alpha\\(1,4-\cos\alpha)^2 + \cos^2\alpha = 1\end{cases}$
$\Leftrightarrow \begin{cases}\sin\alpha = 1,4 -\cos\alpha\\ 2\cos^2\alpha - 2,8\cos\alpha + 0,96= 0\end{cases}$
$\Leftrightarrow \begin{cases}\sin\alpha = 1,4 -\cos\alpha\\\left[\begin{array}{l}\cos\alpha = 0,6\\\cos\alpha = 0,8\end{array}\right.\end{cases}$
$\Leftrightarrow \left[\begin{array}{l}\begin{cases}\sin\alpha = 0,8\\\cos\alpha = 0,6\end{cases}\\\begin{cases}\sin\alpha = 0,6\\\cos\alpha = 0,8\end{cases}\end{array}\right.$
b) Ta có:
$\quad \sin\alpha +\cos\alpha = m$
$\Rightarrow (\sin\alpha +\cos\alpha)= m^2$
$\Rightarrow 1 + 2\sin\alpha\cos\alpha = m^2$
$\Rightarrow \sin\alpha\cos\alpha =\dfrac{m^2-1}{2}$
Khi đó:
$\quad \sin^3\alpha +\cos^3\alpha$
$= (\sin\alpha +\cos\alpha)^3 - 3\sin\alpha\cos\alpha(\sin\alpha +\cos\alpha)$
$= m^3 - \dfrac{3(m^2-1)m}{2}$
$= \dfrac{3m-m^3}{2}$
