Posted by **Natash** on Saturday, November 22, 2008 at 9:08pm.

Find all solutions of the equation 2sin^2x-cosx=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)(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

- trig -
**drwls**, Saturday, November 22, 2008 at 9:37pm
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.

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