If 4 sinα cosβ+2cosβ−3=23sinα . where α,β∈[0,2π] then minimum of α−β=

2cosβ(2sinα+1)−3(2sinα+1)=a ∴(2sinα+1)(2cosβ−3)=0∴sinα=12 (or) cosβ=32α=7π6,11π6 β=π6,11π67π6−11π6=−4π6=−2π3

If 4 sinα cosβ+2cosβ−3=23sinα . where α,β∈[0,2π] then minimum of α−β=

2cosβ(2sinα+1)−3(2sinα+1)=a ∴(2sinα+1)(2cosβ−3)=0∴sinα=12 (or) cosβ=32α=7π6,11π6 β=π6,11π67π6−11π6=−4π6=−2π3

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If $4 \sin α \cos β + 2 \cos β - \sqrt{3} = 2 \sqrt{3} \sin α$ . where $α, β \in [0, 2 π]$ then minimum of $α - β =$

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$- 2 π$

$- \frac{2 π}{3}$

$\frac{- 4 π}{3}$

$- π$

answer is A.

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Detailed Solution

$2 \cos β (2 \sin α + 1) - \sqrt{3} (2 \sin α + 1) = a \begin{array}{l} ∴ (2 \sin α + 1) (2 \cos β - \sqrt{3}) = 0 \\ ∴ \sin α = \frac{1}{2} (o r) \cos β = \frac{\sqrt{3}}{2} \\ α = \frac{7 π}{6}, \frac{11 π}{6} β = \frac{π}{6}, \frac{11 π}{6} \\ \frac{7 π}{6} - \frac{11 π}{6} \\ = - \frac{4 π}{6} = - \frac{2 π}{3} \end{array}$