Đáp án:
$\begin{array}{l}
f){\sin ^2}x + {\sin ^2}2x = {\sin ^2}3x\\
\Rightarrow {\sin ^2}3x - {\sin ^2}x = {\sin ^2}2x\\
\Rightarrow \left( {\sin 3x + \sin x} \right)\left( {\sin 3x - \sin x} \right) = {\sin ^2}2x\\
\Rightarrow 2.\sin 2x.{\mathop{\rm cosx}\nolimits} .\left( { - 2} \right).\cos 2x.\sin x = {\sin ^2}2x\\
\Rightarrow - 2.\sin 2x.\cos 2x.\left( {2.\sin x.\cos x} \right) = {\sin ^2}2x\\
\Rightarrow {\sin ^2}2x + \sin 4x.\sin 2x = 0\\
\Rightarrow \sin 2x.\left( {\sin 2x + \sin 4x} \right) = 0\\
\Rightarrow \sin 2x.2.sin3x.cosx = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 3x = 0\\
\cos x = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = k\pi \\
3x = k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{6}\\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Rightarrow x = \dfrac{{k\pi }}{6}\\
5)a)Dkxd:y = \dfrac{{1 + \sin x}}{{\cos 3x}}\\
\Rightarrow \cos 3x \ne 0\\
\Rightarrow 3x \ne \dfrac{\pi }{2} + k\pi \\
\Rightarrow x \ne \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}
\end{array}$
