kalamandbDmi
Answered

for the first derivative of f(theta)=2cos(theta)+cos^2(theta),

I got

f'(theta)=-2sin(theta)+2cos(theta)-sin(theta)=0

After simplifying it, the book had

(-sin(theta))(1+cos(theta))=0 ... how did they get (1+cos(theta))??? the answers is prob really simple but i am stuck on it if anyone could kindly help. @Mathematics

Verified Answer :

caylus

Hello,

f(t)=2cos t +cos² t

f'(t)=-2sin t -2cos tsin t
     =-2sin(t)(1+cos (t))
Maybe we must calculate f'(t)=0 ==>sin(t)(1+cos(t))=0

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