Đáp án:
$\begin{array}{l}1) \,\left[\begin{array}{l}x = k\pi\\x =- \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\2)\, \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x =\dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)\end{array}$
Giải thích các bước giải:
$\begin{array}{l}1) \,\,(1 - \tan x)(1 + \sin2x) = 1 + \tan x\\ ĐKXĐ:\, \cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{2} +n\pi \quad (n \in \Bbb Z)\\ \text{Nhận thấy $\tan x = 1$ không là nghiệm của phương trình}\\ \text{Chia 2 vế của phương trình cho $1 - \tan x$, ta được:}\\ 1 + \sin2x = \dfrac{1 + \tan x}{1 - \tan x}\\ \Leftrightarrow \sin^2x + 2\sin x\cos x + \cos^2x = \dfrac{\tan x + \tan\dfrac{\pi}{4}}{1 - \tan x.\tan\dfrac{\pi}{4}}\\ \Leftrightarrow (\sin x + \cos x)^2 = \tan\left(x + \dfrac{\pi}{4}\right)\\ \Leftrightarrow (\sin x + \cos x)^2 = \dfrac{\sin\left(x + \dfrac{\pi}{4}\right)}{\cos\left(x + \dfrac{\pi}{4}\right)}\\\Leftrightarrow (\sin x + \cos x)^2 = \dfrac{\sin x + \cos x}{\cos x - \sin x}\\ \Leftrightarrow (\sin x + \cos x)\left(\sin x + \cos x + \dfrac{1}{\sin x - \cos x}\right) = 0\\ \Leftrightarrow (\sin x + \cos x)(\sin^2x - \cos^2x + 1) = 0\\ \Leftrightarrow 2\sin^2x(\sin x + \cos x) =0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin x + \cos x = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin\left(x + \dfrac{\pi}{4}\right) = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x + \dfrac{\pi}{4} = k\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x =- \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ 2)\,\,\cos^2x - \sin^2x = \sin5x + \sin x\\ \Leftrightarrow \cos2x = 2\sin3x\cos2x\\ \Leftrightarrow \cos2x( 1 - 2\sin3x) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos2x =0\\\sin3x =\dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} + k\pi\\3x =\dfrac{\pi}{6} + k2\pi\\3x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x =\dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
