Đáp án:
$\begin{array}{l}
a)Dat:\sin x + \cos x = t\\
\Rightarrow {\left( {\sin x + \cos x} \right)^2} = {t^2}\\
\Rightarrow {\sin ^2} + 2.\sin x.\cos x + {\cos ^2}x = {t^2}\\
\Rightarrow 1 + 2.\sin x.\cos x = {t^2}\\
\Rightarrow 4\sin x.\cos x = 2{t^2} - 2\\
Pt:\sin x + \cos x - 4\sin x.\cos x + 4 = 0\\
\Rightarrow t - \left( {2{t^2} - 2} \right) + 4 = 0\\
\Rightarrow t - 2{t^2} + 2 + 4 = 0\\
\Rightarrow 2{t^2} - t - 6 = 0\\
\Rightarrow 2{t^2} - 4t + 3t - 6 = 0\\
\Rightarrow \left( {t - 2} \right)\left( {2t + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
t = 2\\
t = - \dfrac{3}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2.\sin x.\cos x = {t^2} - 1 = 3\\
2.\sin x.\cos x = {t^2} - 1 = \dfrac{5}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sin 2x = 3\left( {ktm} \right)\\
\sin 2x = \dfrac{5}{4}\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow pt\,vo\,nghiem\\
b)\left\{ \begin{array}{l}
\sin x + \cos x = t\\
2\sin x.\cos x = \sin 2x = {t^2} - 1
\end{array} \right.\\
\sin x + \cos x + 5.\sin 2x + 6 = 0\\
\Rightarrow t + 5\left( {{t^2} - 1} \right) + 6 = 0\\
\Rightarrow 5{t^2} + t + 1 = 0\left( {vo\,nghiem} \right)\\
Vay\,pt\,vo\,nghiem
\end{array}$
