Question

e^(2sinx)=1 ; 0<=x<=16

Answer 1

the only way for
\[e^y=0\] is if
\[y=0\] so you should solve
\[2\sin(x)=0\] or
\[\sin(x)=0\]

Answer 2

this one was pretty :( you stole it satellite

Answer 3

lots of choices on the interval [0,16] but i will let myininaya write them for you

Answer 4

take the ln of both sides

\[\ln (e ^{2Sinx})=\ln(1)\]

Answer 5

5pi is approximately 16
so we are basically looking at the interval [0,5pi]

Answer 6

then for logarithmiic rules you get
\[2\sin (x)=0\]

Answer 7

sinx is 0 when
x=0,pi,2pi,3pi,4pi,5pi

Answer 8

remember 5pi<16 so
we include 5pi

Answer 9

2sin(x)=0 this equation is cero when x=0 or x=180

son the only answers would be 0

if the parameter is in radians the woudl be 0, pi, 2pi , 3pie 4pie ..............5,16pie

Answer 10

can i have some pie?