Đáp án:

$S =\left\{\dfrac{\pi}{6} + k2\pi;\ \dfrac{5\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}$

Giải thích các bước giải:

$\quad \dfrac{1 +\cos2x}{\cos x}=\dfrac{\sin2x}{1 -\cos2x}\qquad (*)$

$ĐK:\begin{cases}\cos x \ne 0\\\cos2x \ne 1\end{cases}\Leftrightarrow \begin{cases}x \ne \dfrac{\pi}{2} + n\pi\\x \ne n\pi\end{cases}\Leftrightarrow x \ne \dfrac{n\pi}{2}$

$(*)\Leftrightarrow (1+\cos2x)(1-\cos2x) = \sin2x.\cos x$

$\Leftrightarrow 1-\cos^22x - \sin2x.\cos x = 0$

$\Leftrightarrow \sin^22x -\sin2x.\cos x = 0$

$\Leftrightarrow \sin2x(\sin2x -\cos x)= 0$

$\Leftrightarrow 2\sin x\cos^2x(2\sin x -1)= 0$

$\Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin x =\dfrac12\end{array}\right.\quad (Do\ \cos x \ne 0)$

$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

Kết hợp ĐKXĐ, ta được:

$\left[\begin{array}{l}x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

Vậy $S =\left\{\dfrac{\pi}{6} + k2\pi;\ \dfrac{5\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}$

Đáp án:

$S=\left\{\dfrac{\pi}{6}+k2\pi;\dfrac{5\pi}{6}+k2\pi\,\bigg|\,k\in\mathbb Z\right\}$

Giải thích các bước giải:

ĐKXĐ: $\begin{cases}\cos x\ne 0\\\cos2x\ne 1\end{cases}⇔\begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\x\ne k\pi\end{cases}$

$\dfrac{1+\cos2x}{\cos x}=\dfrac{\sin2x}{1-\cos2x}$

$⇔(1+\cos2x)(1-\cos2x)=\sin2x.\cos x$

$⇔1-\cos^22x=\sin2x.\cos x$

$⇔\sin^22x-\sin2x.\cos x=0$

$⇔\sin2x(\sin2x-\cos x)=0$

$⇔\sin2x(2\sin x\cos x-\cos x)=0$

$⇔\sin2x.\cos x.(2\sin x-1)=0$

$⇔\left[ \begin{array}{l}\sin2x=0\\\cos x=0\\\sin x=\dfrac{1}{2}\end{array} \right.$

$⇔\left[ \begin{array}{l}2x=k\pi\\x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$

$⇔\left[ \begin{array}{l}x=k\dfrac{\pi}{2}\,(L)\\x=\dfrac{\pi}{2}+k\pi\,(L)\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$

Vậy $S=\left\{\dfrac{\pi}{6}+k2\pi;\dfrac{5\pi}{6}+k2\pi\,\bigg|\,k\in\mathbb Z\right\}$.