Question

Solve each equation in the interval [0,2pi) rounded to two decimal points
a) (2cos(theta)-1)(sin(theta)-1)=0

Answer 1

em I got
2cos(theta)sin(theta)-2cos(theta)-sin(theta)+1=0
yeah?

Answer 2

If a*b=0, then either a=0 or b=0 or both=0.
This means you have:
\[2\cos(\theta)-1=0 \text{ or } \sin(\theta)-1=0\]

Answer 3

isolate the trig function

Answer 4

\[\cos(\theta)=\frac{1}{2} \text{ or } \sin(\theta)=1 \]

Answer 5

try to solve both of those equations

Answer 6

cos(theta) = 1.047
sin(theta) = 1.57

Answer 7

oh you mean:
\[\theta \approx 1.05 \text{ you should also be able to obtain } \theta \approx 5.24 \text{ from } \cos(\theta)=\frac{1}{2}\]

and \[\theta \approx 1.57 \text{ from the other equation } \sin(\theta)=1 \]

Answer 8

Yes, that is what I meant :) how did you get 5.24?

Answer 9

\[\cos(\theta)=\cos(-\theta)=\cos(-\theta+2 \pi ) \\ \cos(1.05) \approx \cos(-1.05+2\pi) \approx \cos(-1.05+2(3.14)) \\ \cos(1.05) \approx \cos(-1.05+6.28)=\cos(5.23) \\ \text{ but honestly I used the unit circle }\]

Answer 10

https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/720px-Unit_circle_angles_color.svg.png

Answer 11

You see the points on the circle.
The first number of every pair is the output for the cos(the angle there).
The second number of every pair is the output for the sin(the angle there).
I was looking for when the x-coordinate (the first number aka the cos number) was 1/2. I see this was happening at the pi/3 mark and the 5pi/3 mark.

Answer 12

And your question wanted these values rounded to the nearest hundredth.

Answer 13

\[\frac{\pi}{3} \approx 1.05 \\ \frac{5 \pi}{3} \approx 5.24\]

Answer 14

ooo I see! that makes sense! haha why do we not want to do the othe pi/3 s?

Answer 15

at 2pi/3 and 4pi/3
you should see the cos number (the x-coordinate) is negative

Answer 16

1/2 is not a negative number

Answer 17

at pi/3 it is not negative either.

That was another question I have
why do we make it
-1.05 instead of adding 2pi to 1.05

Answer 18

cos is an even function
\[\cos(\theta)=\cos(-\theta)\]

Answer 19

sin is an odd function
\[\sin(\theta)=-\sin(-\theta)\]

Answer 20

ooh yes.

Answer 21

and then inside you can add as many 2pi as you like because we just wind up at the same place

Answer 22

but I add one 2pi since that would give me another solution in [0,2pi)

Answer 23

Pretend we want to solve this:
\[\cos(\theta)=.23 , \theta \in [0,2\pi)\]
This .23 is not on the unit circle. But using the unit circle and other identities helps me to solve this equation.

First I can do arccos( ) on both sides to find one solution.
\[\theta=\arccos(.23) \]


another solution would be given by:
\[-\theta+2\pi=\arccos(.23) \\ \theta=-\arccos(.23)+2\pi\]
and if we wanted to approximate these solutions by rounding to the nearest hundredth we would have:
\[\theta \approx 4.94 , 1.34 \]

Answer 24

The second solution I used the fact that cos was even and had a period of 2pi.

Answer 25

Hmm ok.

Answer 26

http://openstudy.com/users/babynini#/updates/55823438e4b07028ea611c18
so for this one, how do I do that?

Answer 27

Thank you for the help on this one, btw. :)

Answer 28

np