Question
Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi)
x1= ?
x2=?
x3=?
x1= ?
x2=?
x3=?
Answers
Damon
2 (1 - cos^2 x) - cos x = 1
2 - 2 cos^2 x -cos x = 1
2 cos^2 x + cos x -1 = 0
let y = cos x
then
2 y^2 +y -1 = 0
(y+1)(2y-1) = 0
y = .5 or y = -1
so
cos x = .5 or cos x = -1
so
x = 60 degrees or 300 degrees
or 180 degrees
2 - 2 cos^2 x -cos x = 1
2 cos^2 x + cos x -1 = 0
let y = cos x
then
2 y^2 +y -1 = 0
(y+1)(2y-1) = 0
y = .5 or y = -1
so
cos x = .5 or cos x = -1
so
x = 60 degrees or 300 degrees
or 180 degrees
drwls
Rewrite as
2(1-cos^2x)-cosx = 1.
Rearrange as
2cos^2x + cos x -1 = 0,
which can be factored to provide:
(2cosx -1)(cosx +1) = 0.
This means
cosx = 1/2 or -1
Use that to find the three angles.
2(1-cos^2x)-cosx = 1.
Rearrange as
2cos^2x + cos x -1 = 0,
which can be factored to provide:
(2cosx -1)(cosx +1) = 0.
This means
cosx = 1/2 or -1
Use that to find the three angles.