Math
posted by Bryan on .
Solve the given equation in the interval [0,2 pi].
Note: The answer must be written as a multiple of pi. Give exact
answers. Do not use decimal numbers. The answer must
an integer or a fraction. Note that pi is already provided with the
answer so you just have to find the appropriate multiple. E.g.
the answer is p/2 you should enter 1/2. If there is more than one
answer write them separated by commas.
2(sin(x))^2−5cos(x)+1 = 0
x=__________ pi

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