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

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.0em I got 2cos(theta)sin(theta)2cos(theta)sin(theta)+1=0 yeah?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2If a*b=0, then either a=0 or b=0 or both=0. This means you have: \[2\cos(\theta)1=0 \text{ or } \sin(\theta)1=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2isolate the trig function

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\cos(\theta)=\frac{1}{2} \text{ or } \sin(\theta)=1 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2try to solve both of those equations

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0cos(theta) = 1.047 sin(theta) = 1.57

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oh 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 \[\theta \approx 1.57 \text{ from the other equation } \sin(\theta)=1 \]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Yes, that is what I meant :) how did you get 5.24?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\cos(\theta)=\cos(\theta)=\cos(\theta+2 \pi ) \\ \cos(1.05) \approx \cos(1.05+2\pi) \approx \cos(1.05+2(3.14)) \\ \cos(1.05) \approx \cos(1.05+6.28)=\cos(5.23) \\ \text{ but honestly I used the unit circle }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2You 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 xcoordinate (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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2And your question wanted these values rounded to the nearest hundredth.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{\pi}{3} \approx 1.05 \\ \frac{5 \pi}{3} \approx 5.24\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0ooo I see! that makes sense! haha why do we not want to do the othe pi/3 s?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2at 2pi/3 and 4pi/3 you should see the cos number (the xcoordinate) is negative

freckles
 one year ago
Best ResponseYou've already chosen the best response.21/2 is not a negative number

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0at 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.2cos is an even function \[\cos(\theta)=\cos(\theta)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2sin is an odd function \[\sin(\theta)=\sin(\theta)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and then inside you can add as many 2pi as you like because we just wind up at the same place

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but I add one 2pi since that would give me another solution in [0,2pi)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Pretend 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: \[\theta \approx 4.94 , 1.34 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2The second solution I used the fact that cos was even and had a period of 2pi.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/users/babynini#/updates/55823438e4b07028ea611c18 so for this one, how do I do that?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for the help on this one, btw. :)
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