a: \(4\cdot\sin^3x+3\cdot\sin^2x\cdot cosx-\sin x-cos^3x=0\)
=>\(4\cdot\sin^3x+3\cdot\sin^2x\cdot cosx-\sin x\left(\sin^2x+cos^2x\right)-cos^3x=0\)
=>\(4\cdot\sin^3x+3\cdot\sin^2x\cdot cosx-\sin^3x-\sin x\cdot cos^2x-cos^3x=0\)
=>\(3\cdot\sin^3x+3\cdot\sin^2x\cdot cosx-\sin x\cdot cos^2x-cos^3x=0\) (1)
Th1: cosx=0
(1) sẽ tương đương: \(3\cdot sin^3x=0\)
=>\(\sin^3x=0\)
=>sin x=0(vô lý)
=>Loại
TH2: cosx<>0
(1) sẽ tương đương: \(3\cdot\left(\frac{\sin x}{cosx}\right)^3+3\cdot\frac{\sin^2x\cdot cosx}{cos^3x}-\frac{\sin x\cdot cos^2x}{cos^3x}-1=0\)
=>\(3\cdot\tan^3x+3\cdot\tan^2x-\tan x-1=0\)
=>(tan x+1)\(\left(3\cdot\tan^2x-1\right)\) =0
=>\(\left[\begin{array}{l}tanx+1=0\\ 3\cdot\tan^2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}tanx=1\\ \tan x=\frac{1}{\sqrt3}\\ \tan x=-\frac{1}{\sqrt3}\end{array}\right.\)
tan x=1
=>\(x=\frac{\pi}{4}+k\pi\)
\(\tan x=\frac{1}{\sqrt3}\)
=>\(x=\frac{\pi}{6}+k\pi\)
\(\tan x=-\frac{1}{\sqrt3}\)
=>\(x=-\frac{\pi}{6}+k\pi\)
