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Damn instructors

Name: Anonymous 2007-03-12 15:16 ID:z9YmYfA2

Okay I have a horrible instructor. I need help with these practice problems

For trigonometry

"Find the general solutions of the eqasions algebraically"
(^2 is squared)
a. ( 4sin^2 x = 2cosx+1
b. sec(4x) =2

Thanks for any help.

Name: Anonymous 2007-03-12 16:18 ID:6Jl/esVB

sec4x=2
sec4=2/x
0.999990480720735= 2/x
0.999990480720735x=2
x=0.499995240360367

That's a bunch of bullcrap. I don't know how to do this.

Name: Anonymous 2007-03-12 17:44 ID:4itA6eU0

>>2
haha oh wow

Name: Anonymous 2007-03-12 20:00 ID:/aXyqijC

>>2
rofl

>>1
Ok, for the first one replace 4*sin^2(x) with 4*(1 - cos^2(x)), then you get 4 - 4cos^2(x) = 2cos(x) + 1, or 4cos^2(x) + 2cos(x) - 3 = 0. Let u = cos(x), then this is 4u^2 + 2u - 3 = 0. Solve for u using the quadratic formula, and then use cos^-1 on those solutions to get solutions for x.

Second one: sec(4x) = 2, so 1/cos(4x) = 2, so cos(4x) = 1/2, so 4x = (pi/3) + 2n*pi or (-pi/3) + 2n*pi, and x = (pi/12) + n/2 * pi or (-pi/12) + n/2 * pi.
(n is any integer)

Name: Anonymous 2007-03-12 20:19 ID:Heaven

(^2 is squared).

Name: Anonymous 2007-03-13 0:04 ID:Heaven

^5 is high five

Name: Anonymous 2007-03-13 1:01 ID:Heaven

(^2 is squared).

Name: Anonymous 2007-03-13 10:25 ID:Heaven

/sci/ is not your answer sheet.

Name: Anonymous 2007-03-14 18:18 ID:sG4wm08w

>>1

Learn how to do this crap yourself.  Otherwise you're just contributing to the willful idiocy us TAs have to fight against when you get to college.

Name: Anonymous 2007-03-15 1:55 ID:o2MuvkO1

>>9

He's trying to learn it.  He said he had a horrible instructor, so he came here for help (which was a mistake on his part).  But you all get off on the fact that you've passed Calc 2, and you want to ensure that as few people as possible attain (and surpass) your mediocrity. That way you can continue lording it over them.

Name: Anonymous 2007-03-17 14:21 ID:mgAwcIVF

>>10
You tell 'im, Joe

Name: Mr. Anonymous 2007-03-18 9:14 ID:f68k/dM+

search wikipedia, there are a set of rules for that (unfortunately i can't think of any right now, it's been a long time since i last solved (or rather attempted to solve) one of them. but what could help you (A would be x in your case):

sin(2A) = 2*sin(A)*cos(A)
cos(2A) = 2^cos(A) - 2^sin(A)
tan(2A) = (2*tan(A))/(1-2^tan(A)

It's basically a riddle game :) Feel free to use all kinds of trickery

Name: Anonymous 2007-03-18 10:15 ID:14kc2Szk

>>4
Has it right.

The first one was a bit tricky, because of the quadratic formula.

The second one was pretty basic, using the definition of secant.

Name: Anonymous 2007-03-22 5:06 ID:+U5q7QAZ

its 4.98715456^3

Name: Anonymous 2007-03-22 16:03 ID:OLRQOeKY

Use the trigometric identities noob.

Don't change these.
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