Dayv O. answered 05/26/21
Caring Super Enthusiastic Knowledgeable Pre-Calculus Tutor
f'(x)=-2sinx-2sin2x is zero whensin2x= -sinx=sin(-x)
again need to solve sin2x=sin(-x)
A.same as 2x= -x+2πk where k=0,+/-1,+/-2,...
B.and is also the same as 2x=π-(-x)+2πk where k=0,+/-1,+/-2,..
in case A. f'(x)=0 when x=2πk/3where k=0,+/-1,+/-2,... k=0,+/-1,+/-2,
these are all potential local min/max, note f''(x)=-2cosx-4cos2x
when k=0, f''<0 so that is a maximum, at k=1 and 2 f'';>0 so those two are minimums.
then is repeats with k=3
In case B, x=3πk where k=0,+/-1,+/-2,... which are repeats of values found in case A.
so min/max at x=2πk/3 where k=0,+/-1,+/-2,..., maximums at k=0,+/-3,+/-6 minimums at the other points
value of f(x) maximum is 3, minimum of f(x) is -1.5