Wednesday

April 23, 2014

April 23, 2014

Posted by **-** on Sunday, November 15, 2009 at 12:40am.

Interval: [0, 2 pi]

Solve 6cos^2x + 5cosx - 6 = 0

- math -
**drwls**, Sunday, November 15, 2009 at 1:35amFor the first one, sin2x = 1/3

2x = 0.33984, 2.80176 or 6.62302 radians (and one more less than 4 pi)

x = 0.16992, 1.4088, or 3.3115

(and one more less than 2 pi that I will leave you to figure out)

For the second question, factor into

(2cosx +3)(3cosx -2) = 0

Then set each factor = 0 and solve for cos x.

**Related Questions**

Advanced Math - Solve 4sin^2 x + 4sqrt(2) cos x-6 = 0 for all real values of x. ...

Pre Cal - Solve 4sin^2 x + 4sqrt(2) cos x-6 = 0 for all real values of x. My ...

Trig - Compute inverse functions to four significant digits. cos^2x=3-5cosx ...

Pre Cal. - So I posted these...but no one answered them. Are they right? Solve ...

math - Can someone help me solve this problem: cos^2x + 5cosx - 2= 0 I don't ...

advanced functions - Solve 2tan^2x + 1 = 0 Interval : [0, 2 pi]

Pre Calculus - Can someone help me solve this problem: cos^2x + 5cosx - 2= 0 I ...

trig question - Compute inverse functions to four significant digits. cos^2x=3-...

Pre Calculus Repost Please Help - Can someone help me solve this problem: cos^2x...

trig integration - i'm having trouble evaluating the integral at pi/2 and 0. i ...