Given curve, f(x)=2cosx−sin2x f′(x)=−2sinx−2cos2x To find turning points, set f′(x)=0 ⇒−2sinx−2cos2x=0⇒sinx−cos2x=0 ⇒sinx+1−2sin2x=0⇒2sin2x−sinx−1=0 Let u=sinx, then 2u2−u−1=0⇒(2u+1)(u−1)=0 ⇒u=−
1
2
or u=1 ∴sinx=
−1
2
or sinx=1 But x∈[−π,π] So, sinx=
−1
2
⇒x=−
5π
6
,−
π
6
And sinx=1⇒x=
π
2
Thus, there are three solutions for x∈[−π,π] ∴ There are 3 turning points.
