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

Solve 3sin2x - 1 = 0

Interval: [0, 2 pi]

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

- math -
**drwls**, Sunday, November 15, 2009 at 1:35am
For 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.

## Answer this Question

## Related Questions

- Trig - Compute inverse functions to four significant digits. cos^2x=3-5cosx ...
- math trig - Solve for x if 3sin2x - 3sinx = 2 - 4cosx 0<x<180
- Advanced Math - Solve 4sin^2 x + 4sqrt(2) cos x-6 = 0 for all real values of x. ...
- math - Can someone help me solve this problem: cos^2x + 5cosx - 2= 0 I don't ...
- Pre Cal - Solve 4sin^2 x + 4sqrt(2) cos x-6 = 0 for all real values of x. My ...
- Pre Cal. - So I posted these...but no one answered them. Are they right? Solve ...
- trig - Solve for all real values of x 3cos2x - 5cosx = 1
- Solve Inequality - Solve the inequality. Express your solution using interval ...
- Trigonometry - How do I solve these? 1) 2sinxcosx-cosx=0 2) cos^2(x)-0.5cosx=0 3...
- Pre Calculus - Can someone help me solve this problem: cos^2x + 5cosx - 2= 0 I ...

More Related Questions