Question
If $(2\sin\theta+3\cos\theta)=2,$ and $(3\sin\theta-2\cos\theta)=\pm3.$

Answer

Given: $(2\sin\theta-\cos\theta)=2\dots(\text{i})$
We have $(2\sin\theta+3\cos\theta)^2+(3\sin\theta-2\cos\theta)^2$
$=4\sin^2\theta+9\cos^2\theta+12\sin\theta\cos\theta\\ \ +9\sin^2\theta+4\cos^2\theta-12\sin\theta\cos\theta$
$=4\big(\sin^2\theta+\cos^2\theta\big)+9\big(\sin^2\theta+\cos^2\theta\big)$
$=4+9$
$=13$
I.e., $(2\sin\theta+3\cos\theta)^2+(3\sin\theta-2\cos\theta)^2=13$
$\Rightarrow2^2+(3\sin\theta-2\cos\theta)^2=13$
$\Rightarrow(3\sin\theta-2\cos\theta)^2=13-4$
$\Rightarrow(3\sin\theta-2\cos\theta)^2=9$
$\Rightarrow(3\sin\theta-2\cos\theta)=\pm3$

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