[toán 11]lượng giác

C

cobe_tocnoi

sin^2 (x)+sin^2(3x)-3cos^2(x)=0
<=>(1-cos2x)/2+(1-cos6x)/2-3(1+cos2x)/2=0
<=>1-cos2x+1-cos6x-3-3cos2x =0
<=>1 +4cos2x+cos6x=0
<=>1+4cos2x +4cos^3(2x) -3cos2x =0
<=>4cos^3(2x) +cos2x +1 =0
<=>(2cos2x+1)(cos^2 2x -cos2x+1)=0
Tới đây thì dễ rồi
 
L

lovelycat_handoi95

[TEX]sin^2x+sin^23x-3cos^2x=0 \\ \Leftrightarrow 1-cos2x+1-cos6x-3cos2x-3=0 \\ \Leftrightarrow 4cos2x+4cos^32x-3cos2x+1=0 \\ \Leftrightarrow cos2x+4cos^32x+1=0[/TEX]
 
P

pigkun_2low

\Leftrightarrow (1-cos 2x)/2 + (1-cos6x)/2 - 3(1+cos2x)/2=0
\Leftrightarrow 1 + 4cos2x + cos6x =0
\Leftrightarrow 1 + 4cos2x + 4(cos2x)^3 - 3(cos2x) = 0
..........
\Leftrightarrowcos2x=-1/2
..........
 
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