Question

Solve the equation on the interval [0,2pi).
2sin^2x-3sinx+1=0
(2sinx+1)(sinx+1)
I don't think I did the factoring correctly. When I multiply it out to double check I get 2sin^2x+3sinx+1

Answer 1

you just got the signs reversed

(2 sinx - 1)(sinx - 1)

Answer 2

Notice that the original had a negative term.
Your factored form is all positive, so you know for sure it is not correct.

how about
(2sinx - 1)(sinx - 1) = 0 ? , yes that works
so sinx = 1/2 or sinx = 1

let's do the easy one, sinx x = 1 , so x = π/2

for sinx = 1/2, we know that sine is positive in quads I and II by the CAST rule
so x = π/6 or x = π-π/6 = 5π/6

so x = π/2, π/6, 5π/6