$pt<=>cos^7x+sin^5x+sinxcos^6x+sin^4xcosx=cosx+sinx(1)$
$TH1:cosx=0<=>x=...$
thay vào $(1)$ được
$sinx=0(loại)$
$TH2:cosx\neq 0$
chia hai vế của $(1)$ cho $cos^7x$ ta được
$1+\frac{sin^5x}{cos^7x}+\frac{sinx.cos^6x}{cos^7x}+\frac{sin^4xcosx}{cos^7x}=\frac{cosx}{cos^7x}+\frac{sinx}{cos^7x}$
$<=>1+tan^5x\frac{1}{cos^2x}+tanx+tan^4x\frac{1}{cos^2x}=\frac{1}{cos^6x}+tanx\frac{1}{cos^6x}$
$<=>1+tan^5(1+tan^2x)+tanx+tan^4x(1+tan^2x)=(1+tan^2x)^3+tanx(1+tan^2x)^3$
$<=>2tan^5x+2tan^4x+3tan^3x+3tan^2x=0$
$<=>2tan^4x(tanx+1)+3tan^2x(tanx+1)=0$
$<=>(2tan^4x+3tan^2x)(tanx+1)=0$
$<=>tan^2x(2tan^2x+3)(tanx+1)=0$
$<=>tanx=0$ hoặc $tanx=-1$