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If $\sin \theta+2 \cos \theta=1$, then what is $2 \sin \theta-\cos \theta$ equal to ?
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Let $\sin \theta+2 \cos \theta=1 \quad$....
.... Consider $2 \sin \theta-\cos \theta=\alpha$ (let)
squaring and adding $\sin ^{2} \theta+4 \cos ^{2} \theta+4 \sin \theta \cos \theta+4 \sin ^{2} \theta+\cos ^{2} \theta-$
$4 \sin \theta \cos \theta=1+\alpha^{2}$
$\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+4\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+\alpha^{2}$
$\Rightarrow 1+4=1+\alpha^{2} \Rightarrow \alpha^{2}=4 \Rightarrow \alpha=2$
.... Consider $2 \sin \theta-\cos \theta=\alpha$ (let)
squaring and adding $\sin ^{2} \theta+4 \cos ^{2} \theta+4 \sin \theta \cos \theta+4 \sin ^{2} \theta+\cos ^{2} \theta-$
$4 \sin \theta \cos \theta=1+\alpha^{2}$
$\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+4\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+\alpha^{2}$
$\Rightarrow 1+4=1+\alpha^{2} \Rightarrow \alpha^{2}=4 \Rightarrow \alpha=2$
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