Posted by **Jen** on Monday, August 21, 2006 at 6:26pm.

I need to find the exact solutions on the interval [0,2pi) for:

2sin^2(x/2) - 3sin(x/2) + 1 = 0

I would start:

(2sin(x/2)-1)(sin(x/2)-1) = 0

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

what's next?

Ok, what angle has a sin equal to say 1/2

sin (x/2)=1

arc sin (1) = x/2

PI/2=x/2

solve for x

do the same technique for the other solution.

so the solution is pi and the other solution would be

arc sin (1/2) = x/2

pi/6 = x/2

x= pi/3

is that right?

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