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2015-04-16T12:08:05+05:30

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Our first step is to rearrange the equation and factor.2\sin\alpha\tan\alpha+1=\tan\alpha+2\sin\alpha\\2\sin\alpha\tan\alpha-2\sin\alpha-\tan\alpha+1=0\\2\sin\alpha(\tan\alpha-1)-1(\tan\alpha-1)=0\\(\tan\alpha-1)(2\sin\alpha-1)=0

Next use zero product property and get
\tan\alpha-1=0  or  2\sin\alpha-1
\tan\alpha=1  or  \sin\alpha=\frac{1}{2}

Since we want acute \alpha 
\alpha=\frac{\pi}{4}  or  \alpha=\frac{\pi}{6}

So the solutions are \alpha=\frac{\pi}{6},~\frac{\pi}{4}
2 5 2
2015-04-16T13:13:45+05:30

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2sinαtanα +1= tanα +2sinα
2sinαtanα  -2 sinα +1 - tanα=0
2sinα(tanα -1) -1( tanα -1) =0
(2sinα-1)(tanα -1)=0
either (2sinα-1) =0 or (tanα-1)=0
If 2sinα-1=0      ,  if tanα-1=0
2sinα=1          ,      tanα=1
sinα=1/2        ,    tanα=tanπ/4
sinα= sinπ/6           , α=π/4
α=π/6  ,if α is acute


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