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If $\sin 3 α = 4 \sin α \sin (x + α) \sin (x - α)$ then

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detailed solution

Correct option is A

sin3α=4sinαsin(x+α)sin(x−α)⇒3sinα−4sin3α=4sinα(sin2x−sin2α)⇒3−4sin2α=4(sin2x−sin2α)⇒3=4sin2x⇒sin2x=34⇒sin2x=sin2π3 ⇒x=nπ±π3,n∈z

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If sin3α=4sinα sin(x+α) sin(x−α) then

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If $\sin 3 α = 4 \sin α \sin (x + α) \sin (x - α)$ then