Question

Critical numbers of 2cosx+sin^2(x)

Answer 1

\[2(-sinx)+cosx(0) + 2sinxcosx\]

Answer 2

\[-2sinx+2sinxcosx=0\]

Answer 3

\[cosx=\frac{ 2sinx }{ 2sinx }\]

Answer 4

\[cosx=1\]

Answer 5

or should i factor out 2sinx?

Answer 6

\[2sinx(-1+cosx)=0\]

Answer 7

sinx=0 cosx=1

Answer 8

i think those are the answers. it says "critical numbers" so the answers should be >1

Answer 9

the book answer says n(pi)

Answer 10

u didn't solve
\[ \large \sin x=0 \]
and
\[ \large \cos x=1 \]

Answer 11

sin is 0 at pi and 2pi

Answer 12

cos is 1 at 2pi

Answer 13

not sure how they are doing this :(

Answer 14

sin is also zero at pi

Answer 15

ya so i pick what they have in common?

Answer 16

which is pi?

Answer 17

in general sin x=0 when \(x=n*\pi\)

Answer 18

well cos is -1 at pi

Answer 19

u don't have to satisfy both equations at the same time.

Answer 20

hmm, so, pi(n) n is an integer, how would i explain that?

Answer 21

when u solve
\[ \large \sin x(-1+\cos x)=0 \]
u have
\[ \large \sin x=0\qquad\text{or}\qquad \cos x=1 \]
so
\[ \large x=n\pi\qquad\text{or}\qquad x=2n\pi \]
ok?

Answer 22

ahhh thanks @helder_edwin !!! =)

Answer 23

whenever \(\cos x=1\), \(\sin x=0\). so the solutions of the former are included in the solutions of the later.

that's why your book says \(x=n\pi\) and nothing else.

Answer 24

u r very welcome.