Sin2x*(cotx +tan2x)=4cos^2x

Đáp án:

$\Leftrightarrow\begin{cases}x=\dfrac{\pi}2+k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{cases}$ $(k\in\mathbb Z)$

Giải thích các bước giải:

ĐKXĐ:

\(\left\{ \begin{array}{l} {\mathop{\rm s}\nolimits} {\rm{inx}} \ne 0\\ \cos 2x \ne 0 \end{array} \right.\)

\(\begin{array}{l} \sin 2x.\left( {\cot x + \tan 2x} \right) = 4{\cos ^2}x\\ \Leftrightarrow \sin 2x.\left( {\dfrac{{\cos x}}{{\sin x}} + \dfrac{{\sin 2x}}{{\cos 2x}}} \right) = 4{\cos ^2}x\\ \Leftrightarrow 2\sin x.\cos x.\dfrac{{\cos x}}{{\sin x}} + \dfrac{{{{\sin }^2}2x}}{{\cos 2x}} = 4{\cos ^2}x\\ \Leftrightarrow 2{\cos ^2}x + \dfrac{{{{\left( {2\sin x.\cos x} \right)}^2}}}{{2{{\cos }^2}x - 1}} = 4{\cos ^2}x\\ \Leftrightarrow \dfrac{{4{{\sin }^2}x.{{\cos }^2}x}}{{2{{\cos }^2}x - 1}} = 2{\cos ^2}x\\ \Leftrightarrow \dfrac{{4\left( {1 - {{\cos }^2}x} \right).{{\cos }^2}x}}{{2{{\cos }^2}x - 1}} = 2{\cos ^2}x\\ \Rightarrow 4{\cos ^2}x - 4{\cos ^4}x = 4{\cos ^4}x - 2{\cos ^2}x\\ \Leftrightarrow 8{\cos ^4}x - 6{\cos ^2}x = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = 0\\ \cos x = \pm \dfrac{{\sqrt 3 }}{2} \end{array} \right. \end{array}\)

Vậy $\Leftrightarrow\begin{cases}x=\dfrac{\pi}2+k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{cases}$ $(k\in\mathbb Z)$.