Posted by **Kenz** on Wednesday, May 30, 2012 at 10:13pm.

Find all solutions on the interval (0, 2pi) of the equation: 2(sin^2)t-3sint+1=0

how do you get this one started?

- Math -
**MathMate**, Wednesday, May 30, 2012 at 10:38pm
let s=sin(t), so

2(sin^2)t-3sint+1=0

becomes

2s²-3s+1=0

(2s-1)(s-1)=0

=>

s=1/2 or s=1

=>

sin(x)=1/2 or sin(x)=1

Now solve for x for 0≤x≤2π

- Math -
**Kenz**, Thursday, May 31, 2012 at 9:52pm
That makes so much sense. Thank you!

- Math :) -
**MathMate**, Friday, June 1, 2012 at 7:44am
You're welcome!

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