Posted by Natash on Saturday, November 22, 2008 at 9:08pm.
Find all solutions of the equation 2sin^2xcosx=1 in the interval [0,2pi)
x1= ?
x2=?
x3=?

trig  Damon, Saturday, November 22, 2008 at 9:33pm
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)(2y1) = 0
y = .5 or y = 1
so
cos x = .5 or cos x = 1
so
x = 60 degrees or 300 degrees
or 180 degrees

trig  drwls, Saturday, November 22, 2008 at 9:37pm
Rewrite as
2(1cos^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.
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