phương trình lượng giác

H

hk_latoiday

N

newstarinsky

d) nhân cả 2 vế với $sinx$ ta được
$sinx.cosx.cos2x.cos4x.cos8x=\dfrac{1}{16}sinx\\
\Leftrightarrow \dfrac{1}{2}sin2x.cos2x.cos4x.cos8x=
\dfrac{1}{16}sinx\\
\Leftrightarrow \dfrac{1}{2}sin4x.cos4x.cos8x=\dfrac{1}{8}sinx\\
\Leftrightarrow \dfrac{1}{2}sin8x.cos8x=\dfrac{1}{4}sinx\\
\Leftrightarrow sin16x=sinx$

e)$sin2x.cosx+sinx.cosx=sinx+cos2x+cosx\\
\Leftrightarrow sinx(2cos^2x-1)+cosx(sinx-1)=cos2x\\
\Leftrightarrow sinx.cos2x-cos2x+cosx(sinx-1)=0\\
\Leftrightarrow cos2x(sinx-1)+cosx(sinx-1)=0\\
\Leftrightarrow (sinx-1)(2cos^2x+cosx-1)=0$
 
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N

nhoka3

câu a
$cos^2x+cos^22x+cos^23x+cos^24x=3/2$
\Leftrightarrow $1+cos2x+1+cos6x+1+cos4x+2cos^2{4x}=3$
\Leftrightarrow $2cos4xcos2x+cos4x+2cos^24x=0$
\Leftrightarrow $cos4x(2cos2x+2cos4x+1)=0$
\Leftrightarrow $cos4x(4cos^22x+2cos2x-1)=0$

câu b
$sin^8x+cos^8x=2.(sin^{10}x+cos^{10}x)+\frac{5}4{}cos2x$
\Leftrightarrow $sin^8x(2sin^2x-1)+cos^8x(2cos^2x-1)+\frac{5}4{}cos2x=0$
\Leftrightarrow $cos2x(cos^8x-sin^8x+\frac{5}{4})=0$
+cos2x=0
+$cos^8x-sin^8x+\frac{5}{4}=0$
\Leftrightarrow $(cos^4x-sin^4x)(cos^4x+sin^4x)+\frac{5}{4}=0$
\Leftrightarrow $cos2x(1-\frac{1}{2}sin^22x)+\frac{5}{4}=0$
\Leftrightarrow $2cos2x-cos2x(1-cos^22x)+\frac{5}{2}=0
\Leftrightarrow $-cos^32x+cos2x+\frac{5}{2}=0$
số lẻ wa' hok pjk có sai chỗ nào hok nữa
 
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0

01697981687

câu c:sin^{8}x+cos^{8}x=2(sin^{10}x+cos^{10}x)+\frac{5}{4cos2x}
<=>sin^{8}x(1-2sin^2x)=cos^8x(cos^2x-1)+5/4cos2x
<=>cos2x(sin^8x-cos^8x)=5/4cos2x
<=>-cos^2 2x(1-1/2*sin^2 2x )=5/4cos2x
<=>4cos^2 2x-2cos^2 2x(1-cos^2 2x)+5/2cos2x=0
<=>4cos^5 (2x)+4cos^3( 2x)+5=0
thế la ok
 
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