ya but then cos x - cos x = 0...
(i had actually worked it out using that but i didn't think the answer would be zero... esp since the next problem is the same the only differnce is that it is from pi < x < 3 pi / 2)
i guess i just needed a second untainted opion... thanks
problem: Simplify F(x) = 1 / ( 1+tan^2 x)^.5 - cos x :0<x<pi/2
ok the only thing i can think of to substitute in was (1+tan^2 x)^.5 = sec^2 x ...
so i got the problem down to 1 / (sec^2 x)^.5 + cos and then become stucks
i m thinken i can just say it is...
ok here's the problem i haven't taken trig for like 2 years and suddenly i have been thrown back into it...
problem: 2 cos 2x - 4 cos x + 3 = 0 0 < x < 2 pi
my work (used cos 2 x = 2 cos^2 x - 1)
4 cos^2 x - 2 - 4 cos x + 3 = 0
4 cos^2 x - 4 cos x + 1 = 0
ok now here is my...
Muzza thanks for the response but unfortunatly it made no sense to me...
but i did ask one of my roomates that just walked in the door and he said that if do cos^-1 0 i get 90 so that's pi/2 and then add 180 to 90 gives me 3pi/2 and this make sense to me,
but anyway are these answers...
ok i have been given a number of trig problems, and i have not taken trig for about 2 years, now i can remeber how to do the problems, but for the life of me i can't remeber how to get the answer...
ok here is the problem:
sin 2x + cos x = 0; 0<x<2 pi
here is what i got
sin 2x + cos x...