a.
ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow tanx=\frac{1}{cot2x}\)
\(\Leftrightarrow tanx=tan2x\)
\(\Leftrightarrow2x=x+k\pi\)
\(\Leftrightarrow x=k\pi\) (ktm)
Vậy pt vô nghiệm
b. ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cos2x}{sin2x}=sinx\)
\(\Leftrightarrow\frac{cos2x.cosx+sin2x.sinx}{cosx.sin2x}=sinx\)
\(\Leftrightarrow\frac{cosx}{cosx.sin2x}=sinx\)
\(\Leftrightarrow\frac{1}{sin2x}=sinx\Leftrightarrow sin2x.sinx=1\)
Do \(\left\{{}\begin{matrix}sin2x\le1\\sinx\le1\end{matrix}\right.\) nên \(sin2x.sinx\le1\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sin2x=1\\sinx=1\end{matrix}\right.\) (vô nghiệm)
Vậy pt đã cho vô nghiệm
c.
\(\Leftrightarrow\left[{}\begin{matrix}\pi\left(x+1\right)=\frac{\pi}{3}+k2\pi\\\pi\left(x+1\right)=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\frac{1}{3}+2k\\x+1=-\frac{1}{3}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}+2k\\x=-\frac{4}{3}+2k\end{matrix}\right.\)
