My quesion says:

Solve this equation for 0<x<2(pi)

4sinx cosx -2(square root)3sinx + 2cosx - (square root)3 = 0

Printable View

- February 13th 2008, 03:50 PMonenamelessSin, Cos, Tan
My quesion says:

Solve this equation for 0<x<2(pi)

4sinx cosx -2(square root)3sinx + 2cosx - (square root)3 = 0 - February 13th 2008, 05:03 PMJhevon
- February 14th 2008, 06:42 PMonenameless
Yes, that was what i meant, sorry i could not find out how to type in a square root sign. But thanks, that's pretty much how much i needed.

- February 14th 2008, 07:10 PMJhevon
- February 15th 2008, 02:20 PMonenameless
(2sinx + 1)(2cosx-squareroot3) = 0

2sinx +1 = 0

2sinx = -1

sinx = -0.5

x = -30

2cosx - sr(3) = 0

2cosx = sr(3)

cosx = sr(3) / 2

x = 30 - February 16th 2008, 04:53 PMJhevon
- February 16th 2008, 06:09 PMonenameless
- February 16th 2008, 06:11 PMJhevon
- February 17th 2008, 07:57 AMonenameless
- February 17th 2008, 07:58 PMJhevon
- February 18th 2008, 02:38 PMonenameless
- February 18th 2008, 02:40 PMJhevon