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 $x=color(blue)(pi/4 , (3pi)/4)$