14.How would you solve this?

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Answer 1 · 2017-12-18T13:44:11

#color(blue)(0 . pi/2 , pi/4 , (3pi)/4)#

Let #u=2x#

#3sin(3u)+3sin(u)#

Using identity:

#color(red)(sin(3x)=-sin^3(x)+3cos^2(x)sin(x))#

#3(-sin^3(u)+3cos^2(u)sin(u))+3sin(u)#

Factor:

#3sin(u)(-sin^2(u)+3cos^2(u)+1)=0#

#3sin(u)=0#

#sin(u)=0=>u=0, pi#

But #u=2xcolor(white)(8888)# , so #x=color(blue)(0/2, pi/2)#

#-sin^2(u)+3cos^2(u)+1=0#

Identity:

#color(red)(1-sin^2(u)=cos^2(u))#

#cos^2(u)+3cos^2(u)=0#

#4cos^2(u)=0#

#cos^2(u)=0=>u=pi/2, (3pi)/2#

But #u=2xcolor(white)(8888)#, so #x=color(blue)(pi/4 , (3pi)/4)#