Đáp án:
d. \(\left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:\cos x \ne 0 \to x \ne \dfrac{\pi }{2} + k\pi \\
\dfrac{{2\sin 2x + \sqrt 3 }}{{\cos x}} = 0\\
\to 2\sin 2x + \sqrt 3 = 0\\
\to \sin 2x = - \dfrac{{\sqrt 3 }}{2}\\
\to \left[ \begin{array}{l}
2x = - \dfrac{\pi }{3} + k2\pi \\
2x = \dfrac{{4\pi }}{3} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k\pi \\
x = \dfrac{{2\pi }}{3} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
b.DK:\cos x > 0\\
\dfrac{{\sqrt 2 - 2\cos 3x}}{{\sqrt {\cos x} }} = 0\\
\to \sqrt 2 - 2\cos 3x = 0\\
\to \cos 3x = \dfrac{{\sqrt 2 }}{2}\\
\to \left[ \begin{array}{l}
3x = \dfrac{\pi }{4} + k2\pi \\
3x = - \dfrac{\pi }{4} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
x = - \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)\\
c.DK:\cos x \ne \dfrac{{\sqrt 3 }}{2}\\
\to x \ne \pm \dfrac{\pi }{6} + k2\pi \\
\dfrac{{2\sin x + 1}}{{2\cos x - \sqrt 3 }} = 0\\
\to 2\sin x + 1 = 0\\
\to \sin x = - \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k2\pi \left( l \right)\\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
d.DK:\sin x \ge 0\\
\sqrt {\sin x} .\left( {2\cos x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sin x = 0\\
\cos x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
