SOLUTION: solve 2sin^2x+3sinx-4=0 in the interval 0<x<pi

Question 1035839: solve 2sin^2x+3sinx-4=0 in the interval 0
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
solve 2sin^2x+3sinx-4=0 in the interval 0 ----------
Solve for what?
You should say, "Solve for x."
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2sin^2x+3sinx-4=0
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Sub u for sin(x)
2u^2 + 3u - 4 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=41 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.850781059358212, -2.35078105935821. Here's your graph:

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Ignore the negative value.
sin(x) = ????????? see above
sin(x) = 0.85078...

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