Author

Topic: sec(1+2x)=sqrt(2)

mbarisas
Single Celled Newbie
Member # 26529
Rate Member

posted January 20, 2009 19:27
Hi, I am completely stuck on this inverse trigonometry function. Is anyone willing to help??? ...I definitely need someone to break it down for me in stepbystep format.
I would be oh, so grateful.
the problem is:
sec(1+2x) = the square root of 2
Posts: 1  From: california  Registered: Jan 2009
 IP: Logged


The Famous Druid
Gold Hearted SuperFan!
Member # 1769
Member Rated:

posted January 20, 2009 20:30
Ok, I don't normally do other peoples homework for them, but today I'm in a "YES WE CAN" mood
First, you need to know that...
sec(x) = 1/cos(x) and 1/sqrt(2) = sqrt(1/2)
So your problem becomes... 1/cos(1+2x) = sqrt(2)
which easily becomes
cos(1+2x) = sqrt(1/2)
You should also recognise that cos(pi/4) = sqrt(1/2) (I'm assuming you're working in radians here)
So you're really solving for (1+2x) = pi/4
 If you watch 'The History Of NASA' backwards, it's about a space agency that has no manned spaceflight capability, then does loworbit flights, then lands on the Moon.
Posts: 10680  From: Melbourne, Australia  Registered: Oct 2002
 IP: Logged


TheMoMan
BlabberMouth, a Blabber Odyssey
Member # 1659
Member Rated:

posted January 21, 2009 02:40
___________________ TFD __ Thanks for the reality check, I had forgotten about the COT, SEC,and CSC and had always just worked with the inverse of the three primary functions, cause my calc did not have said buttons. I must have looked at that expression for a minute and had still not figured out the problem until I read your solution.
 Those who would give up essential liberty to purchase a little temporary safety deserve neither liberty nor safety.
Benjamin Franklin,
Posts: 5848  From: Just South of the Huron National Forest, in the water shed of the Rifle River  Registered: Sep 2002
 IP: Logged


spungo
BlabberMouth, a Blabber Odyssey
Member # 1089
Member Rated:

posted January 21, 2009 05:24
quote: Originally posted by The Famous Druid: Ok, I don't normally do other peoples homework for them, but today I'm in a "YES WE CAN" mood
First, you need to know that...
sec(x) = 1/cos(x) and 1/sqrt(2) = sqrt(1/2)
So your problem becomes... 1/cos(1+2x) = sqrt(2)
which easily becomes
cos(1+2x) = sqrt(1/2)
You should also recognise that cos(pi/4) = sqrt(1/2) (I'm assuming you're working in radians here)
So you're really solving for (1+2x) = pi/4
Small correction, old bean 
cos(2*n*pi +/ pi/4) = sqrt(1/2)
(where n is an integer). So,
1 + 2*x = 2*n*pi +/ pi/4
pedantic, I know, but tutors will pick up on that.
Furthermore, as sqrt(1/2) has two values (positive and negative), you could argue that there is another pair of solutions that satisfy the negative sqrt.
 Shameless plug. (Please forgive me.)
Posts: 6529  From: Noba Scoba  Registered: Jan 2002
 IP: Logged


