Đáp án:
\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \arccos \dfrac{{ - 3}}{5} + k2\pi \\
x = - \arccos \dfrac{{ - 3}}{5} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
x = \dfrac{{ - \pi }}{6} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
x = \dfrac{\pi }{6} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
5\cos x + 3 = 0\\
\Leftrightarrow \cos x = - \dfrac{3}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arccos \dfrac{{ - 3}}{5} + k2\pi \\
x = - \arccos \dfrac{{ - 3}}{5} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
DKXD:\,\,\,\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
\sqrt 3 \tan x + 1 = 0\\
\Leftrightarrow \tan x = - \dfrac{1}{{\sqrt 3 }}\\
\Leftrightarrow \tan x = \tan \dfrac{{ - \pi }}{6}\\
\Leftrightarrow x = \dfrac{{ - \pi }}{6} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
DKXD:\,\,\,\sin x \ne 0 \Leftrightarrow x \ne k\pi \,\,\,\left( {k \in Z} \right)\\
\sqrt 3 \cot x - 3 = 0\\
\Leftrightarrow \cot x = \sqrt 3 \\
\Leftrightarrow \cot x = \cot \dfrac{\pi }{6}\\
\Leftrightarrow x = \dfrac{\pi }{6} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
