If 1+sin2α=3sinαcosα, then value of cotα are . . .

Question : If 1+sin²α=3sinαcosα, then values of cotα are

a) -1,1

b) 0,1

c) 1,2

d) -1,-1

Doubt by Zoha

Solution : 

1+sin²α=3sinαcosα

(sin²α+cos²α)+sin²α=3sinαcosα

[∵sin²θ+cos²θ=1]
2sin²α+cos²α=3sinαcosα
2sin²α+cos²α-3sinαcosα=0
2sin²α-3sinαcosα+cos²α=0

By Splitting the Middle Term

2sin²α-(2sinαcosα+1sinαcosα)+cos²α=0
2sin²α-2sinαcosα-1sinαcosα+cos²α=0
2sin²α-2sinαcosα-sinαcosα+cos²α=0
2sinα(sinα-cosα)-cosα(sinα-cosα)=0
(sinα-cosα)(2sinα-cosα)=0

(sinα-cosα) = 0 OR (2sinα-cosα)=0

sinα-cosα = 0
sinα = cosα
1=cosα/sinα
1=cotα
cotα = 1

2sinα-cosα = 0
2sinα = cosα
2=cosα/sinα
2=cotα
cotα = 2

So, cotα = 1 or 2

Hence, c) would be the correct option.