Solve: `sin3alpha=4sinalpha. Sin(theta+alpha).sin(theta-alpha)` where

1.	Solve: `sin3alpha=4sinalpha. Sin(theta+alpha).sin(theta-alpha)` where `alpha ne n pi, n in I`
Answer» `sin3alpha=4sinalpha. Sin(theta+alpha).sin(theta-alpha)` `rArr 3-4sin^(2)alpha=4sin^(2)theta-4sin^(2)alpha` `rArr 4sin^(2)theta=3` `rArr sin^(2)=3/4` `= (sqrt(3)/(2))^(2)=sin^(2)pi/3` `rArr theta=npi+-pi/3`. Ans.

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