Posted by Bryan on Friday, November 30, 2007 at 6:38pm.
2(sin(x))^2−5cos(x)+1 = 0
2(1- cos^2(x))−5cos(x)+1 = 0
2 - 2cos^2(x) - 5cos(x) + 1 = 0
2cos^2(x) + 5cos(x) + 3 = 0
(cosx + 1 )(2cosx + 3) = 0
cosx = -1 or cosx = -3/2
from the first part x = 0 or 2pi
there is no solution to the second part since the cosine function cannot exceed ±1
Thats not correct....
I'm not sure how you got from
2 - 2cos^2(x) - 5cos(x) + 1 = 0
2cos^2(x) + 5cos(x) + 3 = 0
It should be -2cos^2(x) - 5cos(x)+1=0
From there I'm stuck.
sorry. meant
-2cos^2(x) - 5cos(x)+3=0
Oh my, OH my, two errors in the same post, I must be getting careless.
Let's start from
2(1- cos^2(x))−5cos(x)+1 = 0
2 - 2cos^2(x) - 5cos(x) + 1 = 0
2cos^2(x) + 5cos(x) - 3 = 0
(cosx + 3)(2cosx - 1) = 0
cosx = -3 which will not produce an answer
or
cosx = 1/2
so x = pi/3 or 5pi/3 (60º or 300º)
(these answers are correct, I tested them)
Thank you....the pi/3, 5pi/3 is correct.
Thanks for catching my error,
of course if cos x = -1 then x = pi.
I read my own writing as cosx = 1, (let's blame it on eyesight of an old guy, lol)
you said:"I'm not sure how you got from
2 - 2cos^2(x) - 5cos(x) + 1 = 0 "
look at 2(1- cos^2(x)) and expand it, you get
2 - 2cos^x
so x = pi is your only solution