Wednesday

April 16, 2014

April 16, 2014

Posted by **Wendy** on Friday, September 16, 2011 at 4:44pm.

interval [0,2pi)

How do I solve this? Help would be greatly appreciated!

- Trig -
**Damon**, Friday, September 16, 2011 at 5:30pm2(1-cos^2 x) = 2 + cos x

2 - 2 cos^2 x = 2 + cos x

2 cos^2 x + cos x = 0

cos x (2 cos x + 1 ) = 0

cos x = 0 when x = pi/2 or 90 degrees or 3 pi/2 or 270 deg

or

cos x = -1/2 at x = 2pi/3 or 120 degrees or at 4 pi/3 or 240 degrees

**Related Questions**

Math Trig - Find all solutions on the interval [0,2pi] for the following: 2sin^2...

Math - Can I please get some help on these questions: 1. How many solutions does...

trig - I need to find all solutions of the given equations for the indicated ...

Trig - Find all of the solutions between 0 and 2pi: 2sin(x)^2 = 2 + cos(x)

trig - 2sin(x)cos(x)+cos(x)=0 I'm looking for exact value solutions in [0, 3&#...

trig - Find all real numbers in the interval [0,2pi) that satisfy the equation. ...

college - Trig help... equation on the interval [0,2pi]? 2sin^2(x)=sin(x)

trig - slove the equation exactly over the interval [0, 2pi) sinx=1-2sin^2x

math (trig) - Find sin(x/2) if sin(x)= -0.4 and 3pi/2 < or equal to (x) < ...

trig - Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi...