Homework Help: Calculus

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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