My quesion says:

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

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

Printable View

- Feb 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 - Feb 13th 2008, 05:03 PMJhevon
what you typed is confusing...

i suppose you meant $\displaystyle 4 \sin x \cos x - 2 \sqrt{3} \sin x + 2 \cos x - \sqrt{3} = 0$

in that case, factor out the common $\displaystyle 2 \sin x$ from the first two terms:

$\displaystyle \Rightarrow 2 \sin x ( 2 \cos x - \sqrt{3}) + (2 \cos x - \sqrt{3}) = 0$

but now, $\displaystyle 2 \cos x - \sqrt{3}$ is common to both terms, so pull that out

$\displaystyle \Rightarrow (2 \cos x - \sqrt{3})(2 \sin x + 1) = 0$

now continue... - Feb 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.

- Feb 14th 2008, 07:10 PMJhevon
- Feb 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 - Feb 16th 2008, 04:53 PMJhevon
- Feb 16th 2008, 06:09 PMonenameless
- Feb 16th 2008, 06:11 PMJhevon
- Feb 17th 2008, 07:57 AMonenameless
- Feb 17th 2008, 07:58 PMJhevon
- Feb 18th 2008, 02:38 PMonenameless
- Feb 18th 2008, 02:40 PMJhevon