pt siêu khó

Question

$cos^{7}x+sin^{5}x+\frac{1}{2}(cos^{5}x+sin^{3}x)sin2x=cosx+sinx$

never give up · Answer 1 · 8/3/2014 9:48:45 AM

hoặc là thế này :

$pt\Leftrightarrow cos^7x+sin^5x+\frac{1}{2}cos^5xsin2x+\frac{1}{2}sin^3xsin2x=cosx+sinx$

$\Leftrightarrow cos^7x+cos^6xsinx+sin^5x+sin^4xcosx=cosx+sinx$

$\Leftrightarrow cos^6x(cosx+sinx)+sin^4x(sinx+cosx)=cosx+sinx$

$\Leftrightarrow (cosx+sinx)(cos^6x+sin^4x-1)=0$

$\Leftrightarrow (cosx+sinx)(cos^6x+(1-cos^2x)^2-1)=0$

$\Leftrightarrow (cosx+sinx)(cos^6x+cos^4x-2cos^2x)=0$

$\Leftrightarrow (cosx+sinx)cos^2x(cos^2x-1)(cos^2x+2)=0$

Trả lời 03-08-14 09:48 AM

never give up
1

1K 13K

♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ · Answer 2 · 8/3/2014 8:47:47 AM

$pt<=>cos^7x+sin^5x+sinxcos^6x+sin^4xcosx=cosx+sinx(1)$

$TH1:cosx=0<=>x=...$

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

$tanx=-1$

2 Đáp án