Y=2sinx-cos^2(x)

Question

F"(x)=-sinx+cos^2-sin^2(x)

Find the points of inflection and concavities

Answer 1

well, you remember that

inflection is where y" = 0
concave up when y" > 0

For y" I get 2 times your answer, but that doesn't change its properties.
y' = 2cosx(sinx+1)
y" = -sinx + cos^2x - sin^2x
= -sinx + 1 - 2sin^2x
= -(2sinx-1)(sinx+1)

Note that x = 3pi/2 is not a point of inflection, since y' is also zero there.

Now you can clearly see where y"=0 or is positive/negative. Compare that against the graph of y:

http://www.wolframalpha.com/input/?i=+2sinx-cos^2%28x%29