Lời giải:
a)
\(\sin ^23x-\cos ^2x=0\Leftrightarrow (\sin 3x-\cos x)(\sin 3x+\cos x)=0\Rightarrow \left[\begin{matrix} \sin 3x=\cos x\\ \sin 3x=-\cos x\end{matrix}\right.\)
Nếu \(\sin 3x=\cos x=\sin (\frac{\pi}{2}-x)\)
\(\Rightarrow \left[\begin{matrix} 3x=\frac{\pi}{2}-x+2k\pi \\ 3x=\pi -(\frac{\pi}{2}-x)+2k\pi \end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} x=\frac{\pi}{8}+\frac{k}{2}\pi \\ x=\frac{\pi}{4}+k\pi \end{matrix}\right.\)
Nếu \(\sin 3x=-\cos x=\cos (\pi -x)=\sin (x-\frac{\pi}{2})\)
\(\Rightarrow \left[\begin{matrix} 3x=x-\frac{\pi}{2}+2k\pi \\ 3x=\pi -(x-\frac{\pi}{2})+2k\pi \end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} x=-\frac{\pi}{4}+k\pi \\ x=\frac{3}{8}\pi+\frac{k}{2}\pi \end{matrix}\right.\)
b)
\(8\cos ^3x-1=0\Rightarrow \cos x=\frac{1}{2}=\cos (\frac{\pi}{3})\)
\(\Rightarrow \left[\begin{matrix} x=\frac{\pi}{3}+2k\pi \\ x=\frac{-\pi}{3} +2k\pi \end{matrix}\right.\)
c) Dễ thấy \(\tan x, \cot x\neq 0\)
\(\tan x-2\cot x+1=0\Leftrightarrow \tan x-\frac{2}{\tan x}+1=0\)
\(\Leftrightarrow \tan ^2x+\tan x-2=0\)
\(\Leftrightarrow (\tan x+2)(\tan x-1)=0\Rightarrow \left[\begin{matrix} \tan x=-2\\ \tan x=1\end{matrix}\right.\)
Nếu \(\tan x=-2\Rightarrow x=\tan ^{-1}(-2)+k\pi \)
Nếu \(\tan x=1\Rightarrow x=\tan ^{-1}(1)+k\pi =\frac{\pi}{4}+k\pi \)
