We have, f(x)=∣∣
∣∣sinx02cosx0sinx02cosx0sinx∣∣
∣∣
⇒f(x)=sin3x−2cosx(2sinxcosx)
⇒f(x)=sin3x−4sinx(1−sin2x)
⇒f(x)=5sin3x−4sinx
Differentiating w.r.t. x, we get
f′(x)=15sin2xcosx−4cosx
⇒f′(x)=15cosx(sin2x−415)
⇒f′(x)=15cosx(sinx−2√15)(sinx+2√15)
cosx=0 at x=π2
and sinx=2√15 at two points in (0,π)
∴ Total required points =3