Question

Find the extrema of 2sinx + cos2x within the interval [0, 2pi]. Please show detailed steps.

Thank you : )

Answer 1

First you need to compute the derivative.

Answer 2

as Thomas9 said, we find first f'(x). We know that we'll have maximum and minimum points where f'(x)=0 so we'll have extremas at points where f'(x)=0.
f(x)=2sinx+cos2x
f'(x)=2cosx-2sin2x
then we set f'(x)=0 and find the values of x for which f'(x)=0
f'(x)=0
2cosx-2sin2x=0
(2cosx-2sin2x)/2=0/2
cosx-sin2x=0
recall from trigonometric double angle identities that sin2x=2sinxcosx so we'll have
cosx-(2sinxcosx)=0
cosx-2sinxcosx=0
cosx(1-2sinx)=0
cosx=0 1-2sinx=0
x=cos^(-1)0 -2sinx=-1
x=90 degrees sinx=-1/-2
x=pi/2 sinx=1/2
x=sin^(-1)1/2
x=30 degrees
x=pi/6
so the points where the function will have extremas are the points where x=pi/2 and x=pi/6

Answer 3

The part I'm confused about is how you got from cosx - 2sinxcosx = 0 to cosx(1-2sinx)

Answer 4

So it's cos(x)-2sin(x)cos(x)=0
Now you can just factor: cos(x)(1-2sin(x))=0

Just like: x-2xy=x(1-2y)

Answer 5

@johnwessolom88 actually its an algebraic property so if you have ab-ac=a(b-c)
here, both terms have cosx so we can gladly take it out to have
cosx-2sincosx=cosx(1-2sinx)

Answer 6

is just reversing distribution property

Answer 7

just like that

Answer 8

***Facepalm*** duh
Sorry