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Q

quanhaiphu9a

ko hiểu bạn ơi ...................................................................................................
 
T

trantien.hocmai

$(1+2cos3x)sinx+sin2x=(\sqrt{2}sin(2x+\frac{\pi}{4}))^2$
$<->(1+2cos3x)sinx+sin2x=(sin2x+cos2x)^2$
$<->(1+2cos3x)sinx+sin2x=1+2sin2xcos2x$
$<->sinx(1+2(cos3x+cosx)=1+2sin2xcos2x$
$<->sinx+2cos2xsin2x=1+2sin2xcos2x$
 
K

kamenriderw97

[TEX](1 + 2cos3x).sinx + sin2x = 2.sin^2(2x + \pi/4) [/TEX]
[TEX]\Leftrightarrow sinx + sin4x - sin2x + sin2x = 2.(sin2x + cos2x)^2[/TEX]
[TEX]\Leftrightarrow sinx + sin4x = (sin2x)^2 + (cox2x)^2 + sin4x[/TEX]
[TEX]\Leftrightarrow sinx = 1[/TEX]
[TEX]\Leftrightarrow x = \pi/2 + k2\pi[/TEX]
 
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