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, 03:50 PM #1

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- Feb 13th 2008, 05:03 PM #2
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 PM #3

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- Feb 14th 2008, 07:10 PM #4

- Feb 15th 2008, 02:20 PM #5

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- Feb 16th 2008, 04:53 PM #6

- Feb 16th 2008, 06:09 PM #7

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- Feb 16th 2008, 06:11 PM #8

- Feb 17th 2008, 07:57 AM #9

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- Feb 17th 2008, 07:58 PM #10

- Feb 18th 2008, 02:38 PM #11

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- Feb 18th 2008, 02:40 PM #12