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If $f(x)=x^2$ $g(x)=sinx$, $\forall x\in R$, then the set of all $x$ satisfying $fogogof(x)=gogof(x)$ is

$\begin{array}{1 1} x=\pm\sqrt{n\pi}, where n\in \{0,1,2..............\} \\x=\pm\sqrt{n\pi}, where n\in \{1,2..............\} \\ x=\frac{\pi}{2}+2n\pi, \;where\; n\in \{.......-2,-1,0,1,2..............\} \\x=2n\pi,\; where \;n\in \{.......-2,-1,0,1,2..............\} \end{array}$

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$gof(x)=sin(x^2),$ $gogof(x)=sin(sin(x^2))$
$fogogof(x)=\big[sin(sin(x^2))]^2$
Given $fogogof(x)=gogof(x)$
$\Rightarrow\: \big[sin(sin(x^2))\big]^2=sin(sin(x^2))$
$\Rightarrow\: \big[sin(sin(x^2))\big]^2-sin(sin(x^2))=0$
$\Rightarrow\: \big[sin(sin(x^2))\big]\bigg[sin(sin(x^2))-1\bigg]=0$
$\Rightarrow\:sin(sin(x^2))=0\:\:or\:\:sin(sin(x^2))=1$
$\Rightarrow x^2=n\pi$ or $sin(x^2) =(2n+1)\frac{\pi}{2}$
But $sin(sin(x^2))=1$ is not possible.
$\therefore$ $x=\pm\sqrt{n\pi}$ where $ n\in \{0,1,2..............\}$
answered May 7, 2013 by rvidyagovindarajan_1
edited May 17, 2014 by rohanmaheshwari0831_1
 

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