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If $2 \cos ^{2} \theta+3 \cos \theta=2$, then permissible value of $\cos \theta$ is
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$\frac{1}{2}$
(C)
We have $2 \cos ^{2} \theta+3 \cos \theta=2$
$2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0 \Rightarrow 2 \cos \theta(\cos \theta+2)-1(\cos \theta+2)=0$
$(2 \cos \theta-1)(\cos \theta+2)=0$
$\therefore \cos \theta=\frac{1}{2},-2($ Impossible $) \Rightarrow \cos \theta=\frac{1}{2}$
We have $2 \cos ^{2} \theta+3 \cos \theta=2$
$2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0 \Rightarrow 2 \cos \theta(\cos \theta+2)-1(\cos \theta+2)=0$
$(2 \cos \theta-1)(\cos \theta+2)=0$
$\therefore \cos \theta=\frac{1}{2},-2($ Impossible $) \Rightarrow \cos \theta=\frac{1}{2}$
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