1)
i0
y = sin x + .5 sin2x
y' = cos x + cos2x
y' = 2cos^2x + cosx -1
y' = 0
2cos^2x + cosx - 1 = 0 \Rightarrow 2a^2 + a -1 = 0 I let a = cos x, solve quadratic
a = 0.5, a =-1
cos x= 0.5, cosx =-1
x = \Pi /3, 5\Pi /3 , \pi using the ASTC thingo
y = 3\sqrt[]{3}/2, 0, 3\sqrt[]{3}/4
ii)
y''...