[toán 11] bt 2

N

newstarinsky

$3) sinx.cosx(sin^2x-cos^2x)=\dfrac{1}{4}\\
\Leftrightarrow -\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}\\
\Leftrightarrow sin4x=-1$

1) ĐK $sin2x\not=0$
PT tương đương
$sin2x(sin2x+cos2x)=1-cos2x\\
\Leftrightarrow sin^22x-1+sin2x.cos2x+cos2x=0\\
\Leftrightarrow (sin2x-1)(sin2x+1)+cos2x(sin2x+1)=0\\
\Leftrightarrow (sin2x+1)(sin2x+cos2x-1)=0$
 
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G

gau_gau_gau_00

câu 2

( 1+sinx)^2 = cosx
<=>sin+1=cos^2(x)
<=>sin+1+sin^2(x)-1=0
<=>sin^2(x)+sin(x)=0
1/
sin(x)=0 =>x=kpi
2/
sin(x)=-1 =>x=-pi/2 +k2pi
 
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G

gau_gau_gau_00

câu 4

cos^4 x+sin^4 x +cos(x-pi/4).sin(3x-pi/4) -3/2 =0
<=>1-(1/2)sin^2(2x) +(1/2)(sin(4x-pi/2)+sin(2x))-3/2 =0
<=>sin^2(2x)+cos(4x)-sin(2X)+1=0
<=>sin^2(2x)+sin(2x)-2=0
1/
sin(2x)=1
=>x=pi/2 +k2pi
2/
sin(2x)=-2 (loại)
:p;):)>-:D
 
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T

truongduong9083

Chào bạn

Câu 2.
Gợi ý:
Bình phương hai vế ta được:
$$(1+sinx)^4 = cos^2x$$
$$\Leftrightarrow (1+sinx)[(1+sinx)^3-(1-sinx)] = 0$$
Đến đây bạn làm tiếp nhé
 
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