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A window through the walls of our classroom. This is an interactive learning ecology for students and parents in my Pre-Cal Math 40S class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
8 comments:
http://i32.tinypic.com/1z1yaa0.jpg
i was wondering if it was the answers sorry for triple posting
Benofschool,
Good job for trying to look for the answers. Your hard work only shows how dedicated you are to this class and how much effort you give into it.
Unfortunately, i don't think the answers are right.
Let's examine the first question first.
cos(2x)=1/2
Think of this as your regular algebra question. You can see that the only thing that is in your way is the cosine, so what you have to do is use arccos on both side to get rid of the cosine. That will leave you with this:
2x= 1.0472
Then with some more algebra you will get:
x= 0.5236 + kpi
AND the other other angle at quadrant 4 which is 2.6180 +kpi
Another way to solve this is to use the exact values of the unit circle.
cos2x= 1/2
We know that the cosine of 1/2 is pi/3, so that will leave you with this:
2x=pi/3
Now, again solving for x with some algebra we will get this:
x=pi/6+ kpi
AND
the other angle at quadrant 4 which is 5pi/6 +kpi
Im gonna let you finish up the next question. Hopefully that helps you in some way.
-m@rk
When I first looked at that equation, I thought I should "let 2x = Ѳ" which made the equation look more simplified...
cos(2x) = 1/2
Let 2x = Ѳ.
cosѲ = 1/2
Ѳ = π/6 + 2kπ; 5π/6 + 2kπ; k€I.
Then substitute 2x back instead of Ѳ, and solve the rest.
You guys are doing good work on this! A couple of things:
(1) There will not be a question like this on tomorrow's test. (Ah, I heard your sign just there. ;-))
(2) Zepph has half the answer. The rest of it is:
7π/6 + 6kπ, k ε I
and
11π/6 + 6kπ, k ε I
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