We have, f(x)=2‌sin‌x+cos‌2‌x f′(x)=2‌cos‌x−2‌sin‌2‌x f"(x)=−2‌sin‌x−4‌cos‌2‌x For maxima or minima, f′(x)=0 ⇒2‌cos‌x−2‌sin‌2‌x=0 ⇒cos‌x−2‌sin‌x‌cos‌x=0 ⇒cos‌x(1−2‌sin‌x)=0 cos‌x=0 or 1−2‌sin‌x=0 Now,f′′(
Ï€
2
)=−2‌sin‌
Ï€
2
−4‌cos(
2Ï€
2
) =−2+4=2>0 f′′(
Ï€
6
)=−2‌sin‌
Ï€
6
−4‌cos(
2Ï€
6
) =−1−2=−3<0 f′′(
5Ï€
6
)=−2‌sin‌
5Ï€
6
−4‌cos(
10Ï€
6
) =−1−2=−3<0 Hence, fix) is minimum at
Ï€
2
and maximum at
Ï€
6
∴ f(x) has both local minimum and local maximum.