[toán 11] bt B

G

gau_gau_gau_00

câu 1

1) sin 3x -căn3.cos 3x = 2.sin2x
(1/2)sin(3x)-(căn 3 /2)cos(3x)=sin(2x)
<=>cos(pi/3)sin(3x)-sin(pi/3)cos(3x)=sin(2x)
<=>sin(3x-pi/3)=sin(2x)
<=>3x-pi/3 =2x+k2pi =>x=pi/3 +k2pi
hoặc 3x-pi/3=pi-2x+k2pi =>x=4pi/15 +k2pi/5:p;):)>-
 
G

gau_gau_gau_00

câu 3

(1+sin 2x -cos 2x)/( 1+tan^2 x) =sin2x.cosx
dk:cos#0
pt<=> (1+sin 2x -cos 2x)cos^2(x) =sin2x.cosx
<=> (1+sin 2x -cos 2x) =2sin
<=>2sincos+2sin^2(x)-2sin=0
<=>2sincos+2sin(sin-1)=0
1/
sin=0=>x=kpi (n)
2/
(1/căn 2)cos+(1/căn 2)sin=1/căn 2
<=>sin(pi/4+x)=1/căn 2
=>pi/4+x=pi/4 +k2pi =>x=k2pi (n)
hoặc pi/4 +x=3pi/4 +k2pi =>x=pi/2 +k2pi (n)
:p;):)>-
 
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