Đáp án:
f. \(x = k\pi \left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
k.2\cos x + 3\sin x = 2\\
\to \dfrac{2}{{\sqrt {{2^2} + {3^2}} }}\cos x + \dfrac{3}{{\sqrt {{2^2} + {3^2}} }}\sin x = \dfrac{2}{{\sqrt {{2^2} + {3^2}} }}\\
\to \dfrac{2}{{\sqrt {13} }}\cos x + \dfrac{3}{{\sqrt {13} }}\sin x = \dfrac{2}{{\sqrt {13} }}\\
Đặt:\dfrac{2}{{\sqrt {13} }} = \sin a;\dfrac{3}{{\sqrt {13} }} = \cos a\\
Pt \to \sin a.\cos x + \cos a.\sin x = \sin a\\
\to \sin \left( {a + x} \right) = \sin a\\
\to \left[ \begin{array}{l}
a + x = a + k2\pi \\
a + x = \pi - a + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = k2\pi \\
x = \pi - 2a + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
f.2\left( {1 - 2{{\sin }^2}x} \right) + 3{\sin ^2}x = 2\\
\to - 4{\sin ^2}x + 3{\sin ^2}x + 2 = 2\\
\to - {\sin ^2}x = 0\\
\to \sin x = 0\\
\to x = k\pi \left( {k \in Z} \right)
\end{array}\)
