Question 1035839
solve 2sin^2x+3sinx-4=0 in the interval 0<x<pi
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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
*[invoke solve_quadratic_equation 2,3,-4]
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Ignore the negative value.

sin(x) = ????????? see above
sin(x) = 0.85078...