a/
\(\Leftrightarrow\left[{}\begin{matrix}2cosx+\sqrt{2}=0\\cosx-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{\sqrt{2}}{2}\\cosx=2>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)
b/ ĐKXĐ: ...
\(\Leftrightarrow\left[{}\begin{matrix}tanx-\sqrt{3}=0\\1-tanx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
c/ĐKXĐ: ...
\(\Leftrightarrow\left[{}\begin{matrix}cot\frac{x}{3}=1\\cot\frac{x}{2}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+k\pi\\\frac{x}{2}=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3\pi}{4}+k3\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
Lời giải:
a.
$(2\cos x+\sqrt{2})(\cos x-2)=0$
\(\Rightarrow \left[\begin{matrix} 2\cos x+\sqrt{2}=0\\ \cos x-2=0\end{matrix}\right.\)
Nếu $2\cos x+\sqrt{2}=0\Rightarrow \cos x=\frac{-\sqrt{2}}{2}\Rightarrow x=\pm \frac{3\pi}{4}+2k\pi$ với $k$ nguyên
Nếu $\cos x-2=0\Leftrightarrow \cos x=2$ (vô lý vì $\cos x\leq 1$)
b.
PT \(\Rightarrow \left[\begin{matrix} \tan x=\sqrt{3}\\ \tan x=1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{\pi}{3}+k\pi\\ x=\frac{\pi}{4}+k\pi\end{matrix}\right.\) với $k$ nguyên
c.
PT \(\Rightarrow \left[\begin{matrix} \cot \frac{x}{3}=1\\ \cot \frac{x}{2}=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{3}{4}\pi +3k\pi\\ x=\frac{-\pi}{2}+2k\pi \end{matrix}\right.\) với $k$ nguyên.
