Đáp án:
1) \(\left[ \begin{array}{l}
x = - \dfrac{{2\pi }}{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)y = 0\\
\to \sin x + \sqrt 3 \cos x + 3 = 0\\
\to \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = - \dfrac{{\sqrt 3 }}{2}\\
\to \sin x.\cos \dfrac{\pi }{3} + \sin \dfrac{\pi }{3}.\cos x = - \dfrac{{\sqrt 3 }}{2}\\
\to \sin \left( {x + \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 3 }}{2}\\
\to \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = - \dfrac{\pi }{3} + k2\pi \\
x + \dfrac{\pi }{3} = \dfrac{{4\pi }}{3} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{{2\pi }}{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
2)6{\cos ^2}x + {\cos ^2}2x = 0\\
\to 6{\cos ^2}x + {\left( {2{{\cos }^2}x - 1} \right)^2} = 0\\
\to 6{\cos ^2}x + 4{\cos ^4}x - 4{\cos ^2}x + 1 = 0\\
\to 4{\cos ^4}x + 2{\cos ^2}x + 1 = 0\\
Do:4{\cos ^4}x + 2{\cos ^2}x + 1 > 0\forall x\\
\to x \in \emptyset \\
3)3\sin x + 4\cos x - 1 = 0\\
\to 3\sin x + 4\cos x = 1\\
\to \dfrac{3}{5}\sin x + \dfrac{4}{5}\cos x = \dfrac{1}{5}\\
Đặt:\left\{ \begin{array}{l}
\dfrac{3}{5} = \cos a\\
\dfrac{4}{5} = \sin a\\
\dfrac{1}{5} = \sin B
\end{array} \right.\\
Pt \to \sin x.\cos a + \sin a.\cos x = \sin B\\
\to \sin \left( {x + a} \right) = \sin B\\
\to \left[ \begin{array}{l}
x + a = B + k2\pi \\
x + a = \pi - B + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = B - a + k2\pi \\
x = \pi - B - a + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
