SOLUTION: solve algerbraically over the interval (0,2pi) 2sin^2(x)+3=4 I don't know what I'm supposed. I have a million problems to do just like it. I need steps.

Question 317206: solve algerbraically over the interval (0,2pi)
2sin^2(x)+3=4
I don't know what I'm supposed. I have a million problems to do just like it. I need steps.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
solve algerbraically over the interval (0,2pi)
2sin^2(x)+3=4
.
2sin^2(x)=1
sin^2(x)=1/2
take the square root of both sides:
sin(x) =
Normalizing:
sin(x) =
.
Now, using the "unit circle" discover what angle satisfies the above and it would be:
(pi)/4 and 3(pi)/4

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