Lời giải:
ĐKXĐ: $x\neq k\pi +\frac{\pi}{2}$ với $k$ nguyên
Ta có:
$\sin 2x+2\tan x=0$
$\Leftrightarrow 2\sin x\cos x+2.\frac{\sin x}{\cos x}=0$
$\Leftrightarrow \sin x(\cos x+\frac{1}{\cos x})=0$
$\Rightarrow \sin x=0$ hoặc $\cos x+\frac{1}{\cos x}=0$
Nếu $\sin x=0\Leftrightarrow x=k\pi$ với $k$ nguyên
Nếu $\cos x+\frac{1}{\cos x}=0$
$\Rightarrow \frac{\cos ^2x+1}{\cos x}=0$
$\Rightarrow \cos ^2x=-1<0$ (vô lý)
Kết hợp với đkxđ suy ra $x=k\pi$ với $k$ nguyên bất kỳ.
\(sin2x+2tanx=0\)
\(ĐK:cosx\ne0\)
\(\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)
Đặt \(tanx=t\) \(\Rightarrow sin2x=cos^2x.\dfrac{2sinx.cosx}{cos^2x}\)
\(=\dfrac{1}{tan^2x+1}.2tanx=\dfrac{2t}{t^2+1}\)
\(sin2x+2tanx=0\)
\(\Leftrightarrow\dfrac{2t}{t^2+1}+2t=0\)
\(\Leftrightarrow2t+2t^3+2t=0\)
\(\Leftrightarrow2t\left(t^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=\pm\sqrt{2}\end{matrix}\right.\)
`@` \(t=0\) \(\Rightarrow tanx=0\) \(\Leftrightarrow x=k\pi\left(tm\right)\)
`@`\(t=\sqrt{2}\) \(\Rightarrow tanx=\sqrt{2}\)\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) `(ktm)`
`@`\(t=-\sqrt{2}\) \(\Rightarrow tanx=-\sqrt{2}\) \(\Leftrightarrow x=-\dfrac{\pi}{2}+k\pi\left(tm\right)\)
