Đáp án:
\(x \approx \arctan 0,54 + k\pi \,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\eqalign{
& {\sin ^3}x - {\cos ^3}x + \sin x\cos x\left( {\sin x + \cos x} \right) = 0 \cr
& \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {1 + \sin x\cos x} \right) + \sin x\cos x\left( {\sin x + \cos x} \right) = 0 \cr
& \Leftrightarrow \sin x - \cos x + \sin x\cos x\left( {\sin x - \cos x + \sin x + \cos x} \right) = 0 \cr
& \Leftrightarrow \sin x - \cos x + 2{\sin ^2}x\cos x = 0 \cr
& TH1:\,\,\cos x = 0 \Leftrightarrow x = {\pi \over 2} + k\pi \cr
& \Rightarrow \sin x = \pm 1 \Rightarrow \pm 1 = 0\,\,\left( {Vo\,\,nghiem} \right) \cr
& TH2:\,\,\cos x \ne 0 \Leftrightarrow x \ne {\pi \over 2} + k\pi \cr
& \Leftrightarrow \tan x{1 \over {{{\cos }^2}x}} - {1 \over {{{\cos }^2}x}} + 2{\tan ^2}x = 0 \cr
& \Leftrightarrow \tan x\left( {1 + {{\tan }^2}x} \right) - 1 - {\tan ^2}x + 2{\tan ^2}x = 0 \cr
& \Leftrightarrow {\tan ^3}x + {\tan ^2}x + \tan x - 1 = 0 \cr
& \Leftrightarrow \tan x \approx 0,54 \cr
& \Leftrightarrow x \approx \arctan 0,54 + k\pi \,\,\left( {k \in Z} \right)\,\,\left( {tm} \right) \cr} \)
