Question

Solve the equation on the interval
0<(or equal to) theta<2pi
a.) 2sin (theta)+4=5
b.)sin 3(theta)=1
c.) cos (2(theta)= -1/2
d.) 4sin(theta)+2=0

Answer 1

HI!!
one at a time dear

Answer 2

\[0\le \theta<2\pi\]

Answer 3

and oh sorry i just typed up the whole question

Answer 4

it is 4 questions right?

Answer 5

yeah but it was formatted like that so i just wrote the whole thing

Answer 6

\[2\sin (\theta)+4=5\]

Answer 7

first solve for \(\sin(\theta)\) by subtracting 4 and dividing by 2

Answer 8

let me know when you get
\[\sin(\theta)=\frac{1}{2}\]

Answer 9

i got that

Answer 10

then ask yourself if you know a number whose sine is one half

Answer 11

or look on the unit circle

Answer 12

ok isn't that pi/6?

Answer 13

that is one of them yes

Answer 14

then go across the unit circle and see that \(\frac{5\pi}{6}\) also works

Answer 15

ok

Answer 16

others are similar

Answer 17

hm ok ill try the next one on my own then and get back to you if i get stuck, thanks for your help! :)

Answer 18

\[\sin(3\theta)=1\] ok that is a bit different

Answer 19

oh ok sorry carry on

Answer 20

first solve for \(3\theta\) then divide by 3

Answer 21

\[\sin(3\theta)=1\\
3\theta=\frac{\pi}{2}\\
\theta=\frac{\pi}{6}\] is one answer

Answer 22

then go around the circle again
\[\sin(3\theta)=1\\
3\theta =\frac{5\pi}{2}\\
\theta=\frac{5\pi}{6}\]

Answer 23

you got that?
other two are like the first two

Answer 24

where'd you get the 5pi/2 it's not even on the Unit Circle.. unless my eyes are just failing to see it aha

Answer 25

oh wait it's not supposed to be on the unit circle, is it?

Answer 26

it is an angle
the points on the unit circle are the coordinates

Answer 27

the angle can go around and around and around
that is why sine and cosine are periodic

Answer 28

ohh ok