Thread: Finding Max/Mins of a Function

y = 2cosx + cos(2x)

I guess I'm having trouble with the trig on this (i suck at trig).

The 1st derivative would be -2sinx -2sin2x ... then I try to set that equal to zero and ... yeah. =/ I think sin2x = 2(cosx)(sinx) but then I don't know how to solve for that equal to zero. Any help would be appreciated. Thank you guys.

Originally Posted by drain

y = 2cosx + cos(2x)

I guess I'm having trouble with the trig on this (i suck at trig).

.

I assume you want to maximize this on [0,2pi].

y'= -2sin x - 2 sin 2x

Thus,

-2sin x - 2 sin 2x =0
Divide by -2,
sin x + sin 2x = 0
Double angle identity,
sin x + 2 sin x cos x =0
Thus,
sin x (1+2 cos x) =0
Thus,
sin x =0 or (1+2cos x) =0 on [0,2pi]
Now find its zeros

Thanks alot bro!