What's the sum of the roots of sin x (2sin x - 1) = 2sin x - 1? Can someone check my work and help me find out where I'm making mistakes?

sin x (2sin x - 1) = 2sin x - 1
2 sin²x - sin x = 2sin x - 1
2 sin² x -3sin x + 1 = 0
Let sin x be A
2A² - 3A + 1 = 0
(2A - 1)(A - 1) = 0
So, sin x = 1/2 or sin x = 1

Question

What's the sum of the roots of sin x (2sin x - 1) = 2sin x - 1? Can someone check my work and help me find out where I'm making mistakes?

sin x (2sin x - 1) = 2sin x - 1
2 sin²x - sin x = 2sin x - 1
2 sin² x -3sin x + 1 = 0
Let sin x be A
2A² - 3A + 1 = 0
(2A - 1)(A - 1) = 0
So, sin x = 1/2 or sin x = 1

anonymous · Answer

P.S. The interval is $$0 \le x \le pi$$

anonymous · Answer

So I suppose if sin x = 1/2, then x must be pi/6 or (6/5)pi
And if sin x = 1, then x must be pi/2

anonymous · Answer

Anyway, my book says the sum is 3pi/2

kinggeorge · Answer

If $\sin(x)=1/2$, then $$x={\pi \over 6} \quad \text{or}\quad x={5\pi \over 6}$$

kinggeorge · Answer

Not $$6\pi \over 5$$

anonymous · Answer

Oh, I'm so silly. Thanks!

kinggeorge · Answer

You're welcome.