x is between 0 and 360
Exact answers needed
2006-08-30 05:06:48 · 6 answers · asked by sur2124 4 in Science & Mathematics ➔ Mathematics
2cosx^2 - cosx = 0
(cosx)(2cosx - 1) = 0
cos(x) = 0
x = 90 or 270
2cosx - 1 = 0
2cosx = 1
cosx = (1/2)
x = 60 or 300
x = 60°, 90°, 270°, and 300°
2006-08-30 08:17:00 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋
cosx(2cosx-1)=0
cosx=0, 2cosx=1
x=90 cosx=1/2
or270 x= 60 or300
2006-08-30 07:18:10 · answer #2 · answered by marco_syco 2 · 0⤊ 0⤋
x = 60
2006-08-30 05:10:06 · answer #3 · answered by mbledug 3 · 0⤊ 0⤋
you have two option as far as I know:
1)
simpler method would be factor it out! haven't thought of that before!
OR
2)
you can use quadratic equation to solve this.
you make cosx = y --------------(1)
then you have 2y^2 - y = 0--------------(2)
apply quadratic equation to solve for y and use (1) to get the value of X
clear?
2006-08-30 05:14:28 · answer #4 · answered by David F 2 · 0⤊ 0⤋
cos(x)*(2cos(x)-1)=0 ->cos(x)=0 x=pi/2 ,3pi/2
cos(x)=1/2 x=pi/3,5pi/3
2006-08-30 05:15:07 · answer #5 · answered by amin s 2 · 0⤊ 0⤋
BIG Hint: Factor it into cos(x)*(2cos(x)-1) = 0
Doug
2006-08-30 05:15:12 · answer #6 · answered by doug_donaghue 7 · 0⤊ 0⤋