Question

Solve the following trigonometrical equation when the angle α is acute -
2sinαtanα + 1 = tanα + 2sinα

Answer 1

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 $\alpha=\frac{\pi}{6},~\frac{\pi}{4}$

Answer 2

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