f(x)=1+2sin x+3cos ²x[0,2p÷3) find max value of function

Hello Student,

f(x)=1+2sin x+3cos ²x
f(x)=1+2sin x+ 3 (1 - sin2x)
f(x) = -3sin2x + 2 sin x + 4
df(x)/dx = -6sinxcosx + 2cosx = 0
=>sin x = 1/3 or cosx = 0
d/dx(df(x)/dx) = -3cos2x - 2sinx
<0 for sin x = 1/3
>0 for cos x = 0
=>x = sin-1 (1/3) maxima
=>f(x)_{max} is 13/3

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty