Match the following .
Column –I represents trigonometric equation, Column II – represents general solution of the equation and Column III– represents number of solutions of the equation in  
$[0,\pi ]$

Column -I	Column – II	Column –III
(I) ${\sin }^{2}x+{\cos }^{2}3x=1$	(i) $n\pi$	(P) 1
(II) $e^{{\cot }^2\theta }+{\sin }^2\theta -2{\cos }^{2}2\theta +4=4\sin \theta$	(ii) $\frac{n\pi }{4}$	(Q) 2
(III) $a^2\sin x-a\sin 2x+\sin x=0,a\in R$	(iii) $n\pi +\frac{\pi }{4}$	(R) 3
(IV) $\sqrt[3]{2\tan \theta -1}+\sqrt[3]{\tan \theta -1}=1$	(iv) $(4n+1)\frac{\pi }{2}$	(S) 5

Which of the following combination is correct?

(I) ${\sin }^{2}x+{\cos }^{2}3x=1$
 $\Rightarrow {\sin }^{2}x={\sin }^{2}3x\Rightarrow 3x=n\pi \pm x$
$\Rightarrow 4x=n\pi$  or  $2x=n\pi$
 $\Rightarrow x=\frac{n\pi }{4},\frac{n\pi }{2}\Rightarrow x=\frac{n\pi }{4}$
Number of solution in $[0,\pi ]$ =5
(II)  $e^{{\cot }^2\theta }+{\sin }^2\theta -2{\cos }^{2}2\theta +4=4\sin \theta \Rightarrow e^{{\cot }^2\theta }+{(\sin \theta -2)}^2=2{\cos }^{2}2\theta$
LHS  $\ge 2$, RHS $\le 2\Rightarrow$ LHS =RHS =2
$\Rightarrow {\cot }^2\theta =0,{\cos }^{2}2\theta =1$ and  $\sin \theta =1\Rightarrow \theta =(4n+1)\frac{\pi }{2}$
Number solution in   $[0,\pi ]=1$
(III)  $a^2\sin x-a\sin 2x+\sin x=0\Rightarrow \sin x(a^2-2a\cos x+1)=0$
$\Rightarrow \sin x({(a-\cos x)}^2+{\sin }^{2}x)=0\Rightarrow \sin x=0$ and a = cosx
$\Rightarrow x=n\pi$ number of solution is  $[0,\pi ]=2$
(IV)  $\sqrt[3]{2\tan \theta -1}+\sqrt[3]{\tan \theta -1}=1$
 $\Rightarrow \theta =n\pi +\frac{\pi }{4}$
Number of solutions = 1