Giải phương trình lượng giác.

N

newstarinsky

$b) \sqrt{2}cos2x+sin2x.sin(x+\dfrac{3\pi}{4})=2sin(x+\dfrac{\pi}{4})\\
\Leftrightarrow \sqrt{2}(cos^2x-sin^2x)-\dfrac{1}{\sqrt{2}}sin2x.(sinx+cosx)=\sqrt{2}(sinx+cosx)\\
\Leftrightarrow (sinx+cosx)(cosx-sinx-sinx.cosx)=0$
OK nhé

c) ĐK $cosx\not=0$
PT trở thành
$sinx(sinx-\sqrt{3}cos2x)=2cosx(\sqrt{3}sinx.cosx+sin2x.sinx)\\
\Leftrightarrow sinx[sinx-\sqrt{3}cos2x-2cosx(\sqrt{3}cosx+sin2x)]=0\\
\Leftrightarrow sinx(sinx-4\sqrt{3}cos^2x+\sqrt{3}-4sinx.cos^2x)=0\\
\Leftrightarrow sinx[sinx+\sqrt{3}-4cos^2x(sinx+\sqrt{3})]=0\\
\Leftrightarrow sinxx(sinx+\sqrt{3})(1-4cos^2x)=0$
 
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0

01697981687

chào bạn

câu1: sin3x-3sin2x-cos2x+3sinx+3cosx-2=0
<=>2sin2xcosx-6sinxcosx+2sinx-1+3cosx=2cos^2x
<=>4sinxcos^2x-6sinxcosx+2sinx-1+3cosx=2cos^2x
<=>2cos^2x(2sinx-1)-3cosx(2sinx-1)+2sinx-1=0
<=>(2sinx-1)(2cos^2x-3cosx+1)=0
thế là xong
 
0

01697981687

câu 3:
tanx(sinx-căn3cos2x)=căn3sin2x+cosx-cos3x
dk cosx#0
<=>sin^2x-căn3cos2xsinx=căn3sin2xcosx+4cos^2x-4cos^4x
<=>sin^2x-căn3sin3x=4cos^2xsin^2x
<=>sin^2x-căn3(4sin^3x-3sinx)=4(1-sin^2x)sin^2x
<=>4sin^4x+4căn3sin^3x-3sin^2x-3căn3sinx=0
<=>sinx(4sin^3x+4căn3sin^2x-3sinx-3căn3)=0
thế la xong
 
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