My quesion says:

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

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

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- Feb 13th 2008, 04:50 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 - Feb 13th 2008, 06:03 PMJhevon
- Feb 14th 2008, 07:42 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.

- Feb 14th 2008, 08:10 PMJhevon
- Feb 15th 2008, 03:20 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 - Feb 16th 2008, 05:53 PMJhevon
- Feb 16th 2008, 07:09 PMonenameless
- Feb 16th 2008, 07:11 PMJhevon
- Feb 17th 2008, 08:57 AMonenameless
- Feb 17th 2008, 08:58 PMJhevon
- Feb 18th 2008, 03:38 PMonenameless
- Feb 18th 2008, 03:40 PMJhevon